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NCERT Exemplar for Class 11 Maths Chapter 10 - Straight Lines (Book Solutions)

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NCERT Exemplar for Class 11 Maths - Straight Lines - Free PDF Download

Free PDF download of NCERT Exemplar for Class 11 Maths Chapter 10 - Straight Lines solved by expert Maths teachers on Vedantu.com as per NCERT (CBSE) Book guidelines. All Chapter 10 - Straight Lines exercise questions with solutions to help you to revise the complete syllabus and score more marks in your examinations.


NCERT Exemplar for Class 11 Maths Chapter 10 - Straight Lines explains the concept of equations of straight lines in different formats and how to calculate these equations through a graph. This chapter also explains the types of slopes and the intercepts (x-axis and y-axis) and different properties of straight lines followed by exercises related to the same.

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Access NCERT Exemplar Solutions for CBSE Class 11 Mathematics Chapter 10: Straight Lines

Example 1: Find the equation of a line which passes through the point  (2,3)  and makes an angle of  30o  with the positive direction of  x axis.

Ans: Given: Line passes through (2,3) and makes an angle of  30 with the positive direction of  x  axis.

The slope of line  m=tanθ=tan30=13 .

According to the point-slope form, the equation of a line with slope  m and passing through point  (x,y) is  yy=m(xx) .

y3=13(x2) 

x3y+(332)=0 

Therefore, the equation of the line is  x3y+(332)=0 .


Example 2: Find the equation of a line where length of the perpendicular segment from the origin to the line is  4 and the inclination of the perpendicular segment with the positive direction of  x axis is  30o .

Ans: Given: Length of perpendicular segment from origin to line is  4 .

normal form of the line is xcosω+ysinω=p .

Using p=4 and  ω=30 ,

x32+y12=4 

3x+y=8 

Therefore, the equation of the line is 3x+y=8 .


Example 3: Prove that every straight line has an equation of the form  Ax+By+C=0 , where  AB and  C are constants.

Ans: Given: Every straight line has an equation of the form  Ax+By+C=0 .

A straight line either cuts the  y  axis, or is coincident or parallel to it. The equation of straight line cutting the  y  axis can be expressed in the form  y=mx+b;  and if the straight line is parallel or coincident with  y axis, then its equation will be  x=x1,  where  x=0 in the case of coincidence. The previous two equations are of the form given in the question.

Hence verified.


Example 4: Find the equation of the straight line passing through  (1,2) and  perpendicular to the line  x+y+7=0 .

Ans: Given: The straight line passes through  (1,2) and is perpendicular to  x+y+7=0 .

Let the slope of the required line be  m .

The slope of given line  y=(1)x7 is  1 .

As the two lines are perpendicular, the product of their slopes will be equal to  1 .

m×(1)=1 

m=1 and the line passes through  (1,2) .

So, the equation of required line is  y1=1(x2) .

Therefore, the equation of the line is y1=x2 .


Example 5: Find the distance between the lines 3x+4y=9 and  6x+8y=15 .

Ans:Given: 3x+4y=9 and  6x+8y=15 .

The distance between the parallel lines  y=mx+c1  and  y=mx+c2  is  d=|c1c2|1+m2 

The given lines  3x+4y=9  and  3x+4y=152  are parallel lines.

So, the distance between them is  d=|c1c2|1+m2 .

d=|9152|32+42 

 d=310 

Therefore, the distance between the two lines is  310 .


Example 6: Show that the locus of the mid-point of the distance between the axes of the variable line  xcosα+ysinα=p  is  1x2+1y2=4p2  where  p is a constant.

Ans:

Given: xcosα+ysinα=p .

The intercept form of the line  xcosα+ysinα=p is  xpcosα+ypsinα=1  which gives us the coordinates  pcosα, 0 and 0psinα  where the line intersects  x axis and  y axis.

Let the midpoint of the line segment joining  pcosα, 0 and 0psinα be  (h, k)

So,  h=p2cosα and k=p2sinα .

cosα=p2h and sinα=p2k 

Square both sides of equation and add them.

p24h2+p24k2=1 

1h2+1k2=4p2 

Therefore, the locus is  1x2+1y2=4p2 .


Example 7: If the line joining two points  A(2,0)  and  B(3,1) is rotated about  A  in anticlock wise direction through an angle of  15o . Find the equation  of the line in new position.

The slope of the line joining  A(2,0) and  B(3,1) is  1032=1  i.e.  tan45 .

After rotating the line  15 , the slope of the line will become  tan60=3 .


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So, the equation of new line will be  y0=3(x2) .

y3x+23=0 

Therefore, the equation of line is  y3x+23=0 .


Example 8: If the slope of a line passing through the point  A(3,2) is  34 , then find points on the line which are  5  units away from the point  A .

Ans:  Given:Slope of a line passing through  A(3,2) is  34 .

The equation of line passing through  A(3,2)  with slope  34 is  y2=34(x3) .

4y3x+1=0 

(h,k) is a point on line.

So,  4k3h+1=0

k=3h14 .

(h,k) is a point on line such that  (h3)2+(k2)2=25 .

Substituting  k ,

(h3)2+(3h142)2=25

25h2150h175=0

h26h7=0

h=1,7

Substitute these values in  k=3h14 .

k=1,5

Therefore, the points are either  (1,1) or  (7,5) .


Example 9: Find the equation to the straight line passing through the point of intersection of the lines  5x6y1=0 and  3x+2y+5=0  and perpendicular to the line  3x5y+11=0 .

Ans: Given: Line passes through point of intersection of  5x6y1=0 and  3x+2y+5=0 .

The equation of line passing through intersection of  5x6y1=0 and  3x+2y+5=0 is:  5x6y1+k(3x+2y+5)=0 .

Slope of this line is:  (5+3k)6+2k .

Slope of the line  3x5y+11=0 is  35 .

The product of slopes of perpendicular lines will be  1 .

So,  (5+3k)6+2k×35=1 

 k=45 

So, the equation of required line will be  5x6y1+45(3x+2y+5)=0 

i.e.  5x+3y+8=0 .

Therefore, the equation of line is  5x+3y+8=0 .


Example 10: Aray of light coming from the point  (1,2) is reflected at a point  A on the  x axis and then passes through the point  (5,3) . Find the coordinates of the point  A .

Ans: Given: Theray of light comes from (1,2) .

Let the incident ray strike   x axis at  A having coordinates  (x,0) .


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Slope of reflected ray is:  tanθ=35x ...(1) 

Slope of incident ray is  tan(πθ)=2x1  i.e. tanθ=2x1 ...(2) 

From  (1) and  (2) ,

 35x=2x1 

 x=135 

Therefore, the coordinates of  A are  (135,0) .


Example 11: If one diagonal of a square is along the line  8x15y=0 and one of its vertex is at  (1,2) , then find the equation of sides of the square passing through this vertex.

Ans: Given: 8x15y=0 

Let the square be  ABCD  and the coordinates of  D  be  (1,2) .


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BD is along  8x15y=0  and slope is  815 .

Let the slope of  DC  be  m. 

tan45=m8151+8m15 

15+8m=15m8 

m=237 

So, equation of  DC is  y2=237(x1)  i.e.  23x7y9=0 .

The equation of  AD is  y2=723(x1)  i.e.  7x+23y53=0 .

Therefore, the equations are  7x+23y53=0 and  23x7y9=0 .


Example 12: The inclination of the line  xy+3=0 with the positive direction of  x axis is

(A)  45o 

(B)  135o 

(C)  45o 

(D)  135o 

Ans: Given: xy+3=0 

 xy+3=0 can be written as  y=x+3 .

 m=tanθ=1=tan45  i.e.  θ=45 .

Therefore,  θ=45 .

Correct Option: A


Example 13: The two lines  ax+by=c and ax+by=care perpendicular if

(A) aa+bb=0

(B) ab=ba

(C) ab+ab=0

(D) ab+ba=0

Ans: Given: ax+by=c and ax+by=c.

Slope of  ax+by=c is  ab .

And slope of ax+by=cis  ab .

The lines will be perpendicular if  (ab)(ab)=1  i.e. aa+bb=0.

Therefore, aa+bb=0.

Correct Option: A


Example 14: The equation of the line passing through  (1,2) and perpendicular to  x+y+7=0 is

(A)  yx+1=0 

(B)  yx1=0 

(C)  yx+2=0 

(D)  yx2=0 

Ans:

Given: x+y+7=0 and  (1,2) .

Let the slope of line be  m. 

As the line passes through  (1,2) , the equation of line is  y2=m(x1) .

The lines will be perpendicular if  (m)(1)=1  i.e.  m=1 .

So, the equation of line will be  y2=1(x1)  i.e.  yx1=0 .

Therefore,  yx1=0 .

Correct Options : B


Example 15: The distance of the point  P(1,3)  from the line  2y3x=4 is

(A)  13 

(B)  71313 

(C)  13 

(D) None of these

Ans: Given: 2y3x=4 and  P(1,3) .

The distance of  P(1,3)  from  2y3x=4 will be equal to the length of perpendicular from the given point to the line.

 |2(3)3413|=13 .

Therefore, the distance is 13 .

Correct Options: C


Example 16: The coordinates of the foot of the perpendicular from the point  P(2,3)  on the line  x+y11=0 are

(A)  (6,5) 

(B)  (5,6) 

(C)  (5,6) 

(D)  (6,5) 

Ans: Given: x+y11=0 and  P(2,3) .

Let  (h,k)  be coordinates of foot of perpendicular from  P(2,3) on  x+y11=0 .

Slope of perpendicular line =k3h2 .

Slope of  x+y11=0 is  1 .

The lines will be perpendicular if  (k3h2)(1)=1  i.e.  kh=1 ...(1) 

As  (h,k) lies on  x+y11=0

 h+k11=0 ...(2) 

Solving  (1) and  (2) ,

 h=5, k=6 

Therefore,  (5,6) is the foot of perpendicular.

Correct Option: B


Example 17: The intercept cut off by a line from  y axis is twice than that from  x axis, and the line passes through the point  (1,2) . The equation of the line is

(A)  2x+y=4 

(B)  2x+y+4=0 

(C)  2xy=4                                                                                                                    

(D)  2xy+4=0 

Ans:                                                                                                                                         

Given: Line passes through the point  (1,2)  . Let the intercept made by the line be  a on x axis.                

                    

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Then intercept on y axis will be  2a .

So, equation of line is  xa+y2a=1 .

As the line passes through  (1,2) ,                             

1a+22a=1 

a=2 

So, the equation of line will be  x2+y4=1  i.e.  2x+y=4 .

Therefore,  2x+y=4 .

Correct Option : A


Example 18: A line passes through the point  P(1,2)  such that its intercept between the axes is bisected at  P . The equation of the line is 

(A)  x+2y=5 

(B)  xy+1=0 

(C)  x+y3=0 

(D)  2x+y4=0 

Ans: Given: Line passes through the point  (1,2) .

Let the intercept made by the line be  a on x axis and the intercept on y axis be  b .

So, equation of line is  xa+yb=1 .

As it is bisected at  (1,2)

 1=a+02 and 2=0+b2 

 a=2 and b=4 

So, the equation of line will be  x2+y4=1  i.e.  2x+y=4 .

Therefore,  2x+y=4 .

Correct Option: D


Example 19: The reflection of the point  (4,13)  about the line  5x+y+6=0  is 

(A)  (1,14) 

(B)  (3,4) 

(C)  (0,0) 

(D)  (1,2) 

Ans: Given: (4,13) and  5x+y+6=0 .


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Let  (h,k)  be point of reflection of  (4,13) about  5x+y+6=0 .

The midpoint of line joining  (h,k)  and  (4,13) is  (h+42,k132) .

This point is on  5x+y+6=0 .

So,  5(h+42)+(k132)+6=0 

5h+k+19=0 ...(1) 

Slope of line joining  (h,k) and  (4,13)  =k+13h4 .This lines will be perpendicular to the given line if  (5)(k+13h4)=1 .

h5k69=0 ...(2) 

Solving  (1) and  (2) ,

h=1, k=14 

Therefore,  (1,14)  is the required point.

Correct Option: A


Example 20: A point moves such that its distance from the point  (4,0)  is half that of its distance from the line  x=16 . The locus of the point is 

(A)  3x2+4y2=192 

(B)  4x2+3y2=192 

(C)  x2+y2=192 

(D) None of these

Ans: Given: (4,0) and  x=16 .

Let  (h,k)  be coordinates of the moving point.

Given that the point moves in such a way that its distance from (4,0)  is half that of its distance from x=16 .


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So,  (h4)2+k2=12h1612+0 

(h4)2+k2=14(h16)2 

4(h28h+16+k2)=h232h+256 

3h2+4k2=192 

Therefore,  3x2+4y2=192  is the required locus.

Correct Option: A


Short Answer Type Questions

EXERCISE 10.3:

1. Find the equation of a line which passes through the point  (1,2) and cuts off equal intercepts from axes.

Ans: 

Given: Line passes through the point  (1,2) .

Let the intercept made by the line be  a on x axis and the intercept on y axis be  b .

So, equation of line is  xa+yb=1 .

Given that  a=b .

 xa+ya=1 

It passes through  (1,2) , so  1a2a=1 

 1a=1 

 a=1 

So, equation of line will be  x1+y1=1 

 x+y=1 

Therefore, the equation is  x+y+1=0 .


2. Find the equation of a line which passes through the point  (5,2) and perpendicular to the line joining the points  (2,3) and  (3,1) .

Ans: 

Given: (2,3) and  (3,1) .

Slope of line joining  (2,3) and  (3,1) is   1332=4 .

Slope of the line perpendicular to this line is  14=14

The equation of line that passes through  (5,2) is  y2=14(x5) .

4y8=x5 

x4y+3=0 

Therefore, the equation is  x4y+3=0 .


3.Find the angle between the lines  y=(23)(x+5) and  y=(2+3)(x7) .

Ans: 

Given: y=(23)(x+5) and  y=(2+3)(x7) .

Slope of line  y=(23)(x+5) is   23 .

Slope of the line  y=(2+3)(x7) is  2+3

 θ be the angle between two lines.

Using  tanθ=|m1m21+m1m2| ,

 tanθ=|23231+(23)(2+3)| 

 tanθ=|231+43| 

 tanθ=|3| 

 tanθ=3 or 3 

 tanθ=tan60 or tan120 

Therefore,  θ=60 or θ=120 

4.Find the equation of lines which passes through the point  (3,4) and cuts off intercepts from the coordinate axes such that their sum is  14 .

Ans: 

Given: Line passes through the point  (3,4) .

Let the intercept made by the line be  a on x axis and the intercept on y axis be  b .

So, equation of line is  xa+yb=1 .

Given that  a+b=14 .

 b=14a 

So,  xa+y14a=1 

This equation passes through (3,4) .

 3a+414a=1 

 3(14a)+4aa(14a)=1 

 42+a=14aa2 

 a213a+42=0 

 (a6)(a7)=0 

 a=6,7 

b=146,147

b=8,7

So, the equation of line when a=6 and b=8 is  x6+y8=1  i.e. 4x+3y=24.

The equation of line when a=7 and b=7 is   x7+y7=1  i.e. x+y=7.

Therefore, the equations are  x+y=7 and 4x+3y=24.


5. Find the points on the line  x+y=4 which lie at a unit distance from the line  4x+3y=10 .

Ans: 

Given: x+y=4 and  4x+3y=10 .

Let the point  (x1,y1) be the point lying on  x+y=4  i.e.  x1+y1=4 (1) 

The distance of  (x1,y1) from  4x+3y=10

 |4x1+3y11042+32|=1 

 |4x1+3y1105|=1 

4x1+3y110=±5

Take the  +  sign,

4x1+3y110=5

4x1+3y1=15 ...(2)

From (1)

y1=4x1

Put this value in(2)

4x1+3(4x1)=15

x1=3and y1=43=1.

So, the point is  (3,1) .

Take the    sign,

4x1+3y110=5

4x1+3y1=5 ...(3)

From (1)

y1=4x1

Put this value in(3)

4x1+3(4x1)=5

x1=7and y1=4(7)=11.

So, the point is  (7,11) .

Therefore, the points are  (3,1) and  (7,11) .


6. Show that the tangent of an angle between the lines  xa+yb=1  and  xayb=1 is  2aba2b2 .

Ans: 

Given: The tangent of an angle between  xa+yb=1  and  xayb=1 is  2aba2b2 .

Given that  xa+yb=1 ...(1) 

 xayb=1 ...(2) 

Slope of the line  xa+yb=1 is  ba .

Slope of the line  xayb=1 is  ba .

Using  tanθ=|m1m21+m1m2| ,

 tanθ=|baba1+(ba)(ba)| 

 tanθ=|2ba1(ba)2| 

 tanθ=|2aba2b2| 

 tanθ=2aba2b2 

Hence verified.


7. Find the equation of lines passing through  (1,2) and making angle  30o  with  y axis.

Ans: 

Given: lines passing through  (1,2) .

The line makes  30  with  y axis.

So, the angle made with  x axis is  60 .

Slope of line =tan60=3 .

The equation of line that passes through  (1,2)  and slope  3  is:

 yy1=m(xx1) 

 y2=3(x1) 

 y3x+32=0 

Therefore, the equation is  y3x+32=0 .


8. Find the equation of the line passing through the point of intersection of  2x+y=5 and  x+3y+8=0  and parallel to the line  3x+4y=7 .

Ans: 

Given: Line passes through the point of intersection of  2x+y=5 and  x+3y+8=0  and parallel to  3x+4y=7 .

The equation of line that passes through the point of intersection of  2x+y=5 and  x+3y+8=0 is  (2x+y5)+λ(x+3y+8)=0 .

 (2+λ)x+(1+3y)λ5+8λ=0 

Slope of this line is  (2+λ)1+3λ .

Slope of  3x+4y=7 is  34 .

As the two lines are parallel,  (2+λ)1+3λ=34 

 2+λ1+3λ=34 

 8+4λ=3+9λ 

 5λ=5 

 λ=1 

The equation  (2x+y5)+λ(x+3y+8)=0 becomes  3x+4y+3=0 .

Therefore, the equation is  3x+4y+3=0 .


9. For what values of  a  and  b  the intercepts cut off on the coordinate axes by the line ax+by+8=0 are equal in length but opposite in signs to those cut off by the line2x3y+6=0 on the axes.

Ans: 

Given:ax+by+8=0and 2x3y+6=0.

We can write ax+by+8=0 as:

ax+by=8

a8x+b8y=1

x8a+y8b=1

The intercepts are 8a, 8b.

We can write 2x3y+6=0 as:

2x3y=6

2x63y6=1

x3+y2=1

The intercepts are 3, 2.

From the question, 8a=3, 8b=2 i.e. a=83, b=4

Therefore,  a=83 and  b=4 .


10. If the intercept of a line between the coordinate axes is divided by the point (5,4)in the ratio 1:2, then find the equation of the line.

Ans: 

Given:(5, 4)

Assume aand bare intercepts on given line.

Then, the coordinates are A(a,0)and B(0,b).

Using x=m1x2+m2x1m1+m2and y=m1y2+m2y1m1+m2,

5=1×0+2×a1+2

2a=15

a=152

So, A=(152,0).

 And 4=1×b+0×21+2

4=b3

b=12

So, B=(0,12).

The equation of line ABis yy1=y2y1x2x1(xx1).

y0=(1200+152)(x+152)

y=85(x+152)

5y=8x+60

Therefore,8x5y+60=0.


11. Find the equation of a straight line on which length of perpendicular from the origin is four units and the line makes an angle of  120o with the positive direction of  x axis. 

Ans: 

Given: Line makes  120 with the positive direction of  x axis. 

The line makes  120  with the positive direction of  x axis.

So, slope of the line is  tan120=3 .

Equation of the required line is: 

 y=3x+c 

 3x+yc=0

This line is 4 units away from  (0,0) .

 |3(0)+0c|3+1=4 

 |c|=8 

 c=±8 

Therefore, the equation of line is  3x+y±8=0 .


12. Find the equation of one of the sides of an isosceles right angled triangle whose hypotenuse is given by  3x+4y+4=0  and the opposite vertex of the hypotenuse is  (2,2) .

Ans: 

Given: 3x+4y+4=0 

The hypotenuse is along the line  3x+4y+4=0 

Slope of the hypotenuse is  34 .

As the triangle is isosceles right angled triangle, two angles will be equal to  45 .

Let the slope of the line making  45  with the hypotenuse be  m .

 tan45=|m(34)1+m(34)| 

 tan45=|4m+343m| 

 4m+343m=±1 

 4m+3=43m or 4m+3=3m4 

 m=17 or m=7 

Slope of this line is  17 , so the equation of this line is  y2=(1/7)(x2)  i.e.  x7y+12=0 .

The slope of line perpendicular to it is  7

The equation of this line is:  y2=7(x2)  i.e.  7x+y16=0 .

Therefore, the equation of required line is  7x+y16=0 .


13. If the equation of the base of an equilateral triangle is  x+y2  and the vertex is  (2,1) , then find the length of the side of the triangle.

Ans: 

Given:  x+y2 and  (2,1) .

The altitude from the vertex  (2,1) on the base  x+y2  meets at the midpoint of base. 

The length of altitude is the distance of  (2,1)  from x+y2  =|2+(1)212+12|=12 .

The triangle is an equilateral triangle. So,  sin60=Altitudeside of triangle .

length of side of triangle=Altitudesin60=1232=23 .

Therefore, the length of side of the triangle is  23 .


14. A variable line passes through a fixed point  P . The algebraic sum of the perpendiculars drawn from the points  (2,0),(0,2)  and  (1,1)  on the line is zero. Find the coordinates of the point  P .

Ans: 

Given:  (2,0),(0,2)  and  (1,1) 

Assume that the variable line passing through the fixed point  P is  ax+by+c=0 .

Perpendicular distance from the point A(2,0)=2a+0+ca2+b2 

Perpendicular distance from the point B(0,2)=0+2b+ca2+b2 

Perpendicular distance from the point C(1,1)=a+b+ca2+b2 

Given that,  2a+0+ca2+b2+0+2b+ca2+b2+a+b+ca2+b2=0 

 3a+3b+3c=0 or a+b+c=0 

Therefore, the variable line passes through the fixed point  (1,1) .


15. In what direction should a line be drawn through the point  (1,2)  so that its point of intersection with the line  x+y=4  is at a distance  63  from the given point.

Ans: 

Given:  x+y=4 and  (1,2) .

Let the slope of line be  m

It passes through A(1,2) 

 So, the equation of line is  y2=m(x1)  or  mxy+2m=0 ...(1) 

The equation of the given line is  x+y4=0...(2) 

Assume that these lines meet at point  B .

Solving  (1) and  (2)

B(m+2m+1,3m+2m+1) 

Given that  AB=63 

AB2=69 

(m+2m+11)2+(3m+2m+12)2=69 

(1m+1)2+(mm+1)2=23 

1+m2(1+m)2=23 

3+3m2=2+2m2+4m 

m24m+1=0 

m=4±1642=2±3=2+3 or 23 

tanθ=2+3 or 23 

θ=75 or  θ=15 .

Therefore,  θ=75 or θ=15 


16. A straight line moves so that the sum of the reciprocals of its intercepts made on axes is constant. Show that the line passes through a fixed point. 

Ans: 

Given:  1a+1b=  constant

The equation of line in intercept form is  xa+yb=1 .

As  1a+1b=  constant, 

Let  1a+1b  =1k

So,  ka+kb=1 

Thus,  (k,k)  lies on  xa+kb=1 .

Therefore, the line passes through the fixed point  (k,k) .


17. Find the equation of the line which passes through the fixed point  (4,3)   and the portion of the line intercepted between the axes id divided internally in the ratio 5:3 by this point.

Ans: Let AB be a line passing through a point (- 4, 3) and meets x-axis at A(a, 0) and y-axis at B(0, b).


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⇒∴4=5×0+3a5+3  

4=3a8[X=m1x2+m2x1m1+m2,y=m1y2+m2y1m1+m2] 

3a=32                                                                                                              

a=323   and  3=5×b+3×05+3                                                                

3=5×b85b=24b=255  

Intercept form of the line is  x323+y245=1                      

3x32+5y24=1 

9x+20y=969x20y+96=0  

Hence, The required equation is  9x20y+96=0 


18. Find the equation of the lines through the point of intersection of the lines  xy+1=0   and 2x3y+5=0 and whose distance from the point  (3,2)   is  75.  

Ans: Given equation are

xy+1=0(i)  And  2x3y+5=0(ii)  Solving eq. (i) and eq. (ii) 

we get  2x2y+2=0   2x3y+5=0   

y3=0,y=3 

From eq. (i) we have 

x3+1=0x=2 

So,  (2,3)  is the point of intersection of eq.(i) and eq. (ii). 

Let m be the slope of the required line

   Equation of the line is

 y3=m(x2) 

 y3=mx2m 

 mxy+32m=0 

Since, the perpendicular distance from  (3,2) 

to the line is  75  then,

 75=|m(3)2+32mm2+1| 

 4925=(3m2+32m)2m2+1 

 4925=(m+1)2m2+1 

 49m2+49=25m2+50m+25 

 49m225m250m+4925=0 

 24m250m+24=0 

 12m225m+12=0 

 12m216m9m+12=0 

 4m(3m4)3(3m4)=0 

 (3m4)(4m3)=0 

 3m4=0 and 4m3=0 

 3m4=0 and 4m3=0 

 m=43,34 

Equation of the line taking  m=43  is

 y3=43(x2) 

 3y9=4x84x3y+1=0 

Equation of the line taking  m=34  is

 y3=34(x2) 

 4y12=3x63x4y+6=0 

Hence, the required equation are

 4x3y+1=0 and 3x4y+6=0 


19.If the sum of the distance of a moving point in a plane from the axes is 1, then find the locus of the point.

Ans: Let coordinates of a moving point P be  (x,y).  

Given that the sum of the distance from the axes to the point is always 1


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|x| + |y| = 1 

x + y = 1  

xy=1  

x+y=1 

xy = 1 

Hence, these equations gives us the locus of the point P which is a square.


20. P1, P2 are points on either of the two lines  y3|x|=2 at a distance of 5 units from their point of intersection. Find the coordinates of the foot of perpendiculars drawn from P1, P2 on the bisector of the angle between the given lines.


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Ans: Given lines are  y3|x|=2 

y3x=2   if x ≥ 0 …(i)

And  y+3x=2  if x < 0 …(ii)

Slope of eq. (i) is  tanθ=3    θ=60o  

Slope of eq. (ii) is  tanθ=3    θ=120o 

120° Solving eq. (i) and eq. (ii) we get

y3x=2 

y+3x=2 

__________ 

2y=4y=2  

Putting the value of y is eq. (i) we get

x=0  

Point of intersection of line (i) and (ii) is  Q(0,2)  

QO=2  

 In  ΔPEQ,                                                                                                 

cos30o=PQQE32=PQ5 

⇒∴PQ=532 

OP=OQ+PQ  

=2+532  

Hence, the coordinates of the foot of perpendicular  =(0,2+532)  


21.If p is the length of perpendicular from the origin on the line  xa+yb=1  and  a2,p2,b2  are in A.P. then show that  a4+b4=0.  

Ans: Given equation is xa+yb=1

Since, p is the length of perpendicular drawn from the origin to the given line

p=|0a+0b11a2+1b2|

Squaring both sides, we have

p2=11a2+1b2 1a2+1b2=1p2(i) Since a2,p2,b2 are in A.P. 2p2=a2+b2 p2=a2+b221p2=2a2+b2

Putting the value of 1p2 is eq. (i) we get,

1a2+1b2=2a2+b2

a2+b2a2b2=2a2+b2

(a2+b2)2=2a2b2

a4+b4+2a2b2=2a2b2

a4+b4=0

Hence proved.


22. A line cutting off intercept  3  from the  y axis and the tangent at angle to the  x axis is  35 ,  its equation is 

(A)  5y3x+15=0 

(B)  3y5x+15=0 

(C)  5y3x15=0 

(D) None of these

Ans: 

Given: A line cuts off an intercept  3  from the  y axis.

Let the equation of required line be  y=mx+c .

As  c=3 and  m=35 ,

The equation of line is:  y=35x3 

 5y3x+15=0 

Hence, the equation is  5y3x+15=0 .

Correct Option : A


23. Slope of a line which cut off intercepts of equal lengths on the axes is 

(A)  1 

(B)  0 

(C)  2 

(D)  3 

Ans: 

Given: A line cuts off intercepts of equal lengths on the axes.

The equation of required line is xa+ya=1 .

x+y=a 

y=x+a 

Slope =1 

Hence, the slope is  1 .

Correct Option: A


24. The equation of the straight line passing through the point  (3,2)  and perpendicular to the line  y=x  is

(A)  xy=5 

(B)  x+y=5 

(C)  x+y=1 

(D)  xy=1 

Ans: 

Given: (3,2) and  y=x .

Slope of  y=x is  1 .

So, the slope of line perpendicular to it is  1 .

The line passes through  (3,2) .

So, equation of required line is  y2=1(x3)  i.e.  x+y=5 .

Hence, equation is  x+y=5 .

Correct Option: B


25. The equation of the line passing through the point  (1,2) and perpendicular to the line  x+y+1=0  is

(A)  yx+1=0 

(B)  yx1=0 

(C)  yx+2=0 

(D)  yx2=0 

Ans: 

Given: (1,2) and  x+y+1=0 .

Slope of  x+y+1=0 is  1 .

So, the slope of line perpendicular to it is  1 .

The line passes through  (1,2) .

So, equation of required line is  y2=1(x1)  i.e.  yx1=0 .

Hence, equation is  yx1=0 .

Correct Option: B


26. The tangent of angle between the lines whose intercepts on the axes are  a,b and  b,a , respectively, is

(A)  a2b2ab 

(B)  b2a22 

(C)  b2a22ab 

(D) None of these

Ans: 

Given: The intercepts on the axes are  a, b and  b, a .

The intercepts of line are  a and  b .

 So, the line passes through the points  (a,0) and  (0,b) .

Thus, the slope of line is  m1=b00a=ba .

The intercepts of line are  b and  a .

 So, the line passes through the points  (b,0) and  (0,a) .

Thus, the slope of line is  m2=a00b=ab .

Let  θ  be the angle between the lines.

 tanθ=baab1+abba 

 tanθ=b2a2ab2 

 tanθ=b2a22ab 

Hence,  tanθ=b2a22ab .

Correct Option: C


27. If the line  xa+yb  passes through the points  (2,3) and  (4,5) , then  (a,b)  is

(A)  (1,1) 

(B)  (1,1) 

(C)  (1,1) 

(D)  (1,1) 

Ans: 

Given: (2,3) and  (4,5) .

The points  (2,3)  and  (4,5)  lie on  xa+yb=1 .

 2a3b=1 ...(1) 

And  4a5b=1 ...(2) 

Multiply  (1)  by  2  and subtract  (2) from it.

 6b+5b=1 

 1b=1 

 b=1 

Put  b=1  in (1) ,

 2a+3=1 

 a=1 

So,  (a,b)=(1,1) 

Hence,  (a,b)=(1,1) .

Correct Option: D


28. The distance of the point of intersection of the lines  2x3y+5=0  and  3x+4y=0  from the line  5x2y=0  is

(A)  1301729 

(B)  13729 

(C)  1307 

(D) None of these

Ans: 

Given: 2x3y+5=0 and  3x+4y=0 .

Solve the lines  2x3y+5=0 and  3x+4y=0 .

The point of intersection is  (2017,1517) .

The distance of  (2017,1517) from  5x2y=0  is  |5×(2017)2(1517)|25+4 

The distance of  (2017,1517) from  5x2y=0  is  |100173017|29 

The distance of  (2017,1517) from  5x2y=0  is  1301729 

Hence, the distance is  1301729 .

Correct Option: A


29. The equations of the lines which pass through the point  (3,2)  and are inclined at  60 to the line 3x+y=1  is

(A)  y+2=0,3xy233=0 

(B)  x2=0,3xy+2+33=0 

(C)  3xy233=0 

(D) None of these

Ans: 

Given: (3,2) and  3x+y=1 .

The slope of 3x+y=1 is  m1=3

Let the slope of required line which makes  60  with  3x+y=1 be  m .

Then,  tan60=|3m13m| 

 |3m13m|=3 

 3m=33m or3m=3+3m

 m=3 or m=0 

Line passes through (3,2)

So, the equation of line is:  y+2=3(x3)  or  y+2=0 

 3xy233=0 and y+2=0 

Hence, the equations are  y+2=0, 3xy233=0 .

Correct Option: A


30. The equations of the lines passing through the point  (1,0)  and at a distance  32  from the origin, are

(A)  3x+y3=0,3xy3=0 

(B)  3x+y+3=0,3xy+3=0 

(C)  x+3y3=0,x3y3=0 

(D) None of these.

Ans: 

Given: Line passes through  (1,0) .

Let the slope of the required line be  m .

The equation of line passing through  (1,0)  is: y0=m(x1)  i.e.  ymx+m=0 .

The distance of this line from  (0,0) is  32 .

 32=|00+m|1+m2 

 32=|m|1+m2 

 3+3m2=4m2 

 m2=3 

 m=±3 

Hence, equations of lines are:  3x+y3=0 , 3xy3=0 .

Correct Option: A


31. The distance between the lines  y=mx+c1  and  y=mx+c2  is

(A)  c1c2m2+1 

(B)  |c1c2|1+m2 

(C)  c2c11+m2 

(D) 0

Ans: 

Given: y=mx+c1 and  y=mx+c2 .

Let  P(x1,y1)  be a point on y=mx+c1 .

The other line is:  y=mx+c2  i.e.  mxy+c2=0 .

Distance of  P(x1,y1)  from this line is  d=|mx1y1+c2|m2+1 .

 P(x1,y1) lies on  y=mx+c1 ,

 y1=mx1+c1 

 mx1y1=c1 

 d=|c1c2|m2+1 

Hence, distance is  |c1c2|m2+1 .

Correct Option: B


32. The coordinates of the foot of perpendicular from the point  (2,3)  on the line  y=3x+4  is given by

(A)  (3710,110) 

(B)  (110,3710) 

(C)  (1037,10) 

(D)  (23,13) 

Ans: 

Given:  y=3x+4 and  (2,3) .

Let the foot of perpendicular from  P(2,3)  on  3xy+4=0  be M(h,k) .

 M(h,k) lies on  y=3x+4 .

 3hk+4=0 ...(1) 

The slope of the given line is  3  .

Slope of the line joining  P(2,3)  and  M(h,k)  is  13=k3h2  or  h+3k11=0 ...(2) 

Solve  (1) and  (2)

 (h,k)=(110,3710) 

Hence, the coordinates of foot of perpendicular is  (110,3710) .

Correct Option: B


33. If the coordinates of the middle point of the portion of a line intercepted between the coordinate axes is  (3,2) , then the equation of the line will be

(A)  2x+3y=12 

(B)  3x+2y=12 

(C)  4x3y=6 

(D)  5x2y=10 

Ans: 

Given: (3,2) 

As the middle point is  P(3,2) , the line meets the axes at  A(6,0)  and  B(0,4) .

So, the equation of the line using intercept form is:  x6+y4=1  i.e.  2x+3y=12 .

Hence, the equation is  2x+3y=12

Correct Option: A


34. Equation of the line passing through  (1,2)  and parallel to the line  y=3x1  is

(A)  y+2=x+1 

(B)  y+2=3(x+1) 

(C)  y2=3(x1) 

(D)  y2=x1 

Ans: 

Given: (1,2) and  y=3x1 .

The required line is parallel to y=3x1 .

Thus, slope of the line is  3 .

The line also passes through  (1,2) .

So, the equation of line is:  y2=3(x1) 

Hence, the equation is  y2=3(x1)

Correct Option: C


35.Equations of diagonals of the square formed by the lines  x=0,y=0,x=1  and  y=1  are

(A)  y=x,y+x=1 

(B)  y=x,x+y=2 

(C)  2y=x,y+x=13 

(D)  y=2x,y+2x=1 

Ans: 

Given: x=0, y=0, x=1 and  y=1 .

Using the equations  x=0, y=0, x=1 and  y=1 , the vertices of square are  A(0,0), B(1,0), C(1,1), D(0,1) .

The equation of  AC  is:  y0=1010(x0)  i.e.  xy=0 .

The equation of  BD  is:  y0=1001(x1)  i.e.  x+y1=0 .

Hence, the diagonals are y=x, y+x=1 .

Correct Option: A


36. For specifying a straight line, many geometrical parameters should be known?

(A)  1 

(B)  2 

(C)  4 

(D)  3 

Ans: 

Given: A straight line is specified.

The general equation of a linear equation in two variables is  ax+by+c=0 

At least one of  a  and  b  should be non-zero. 

Assume a0

Then, the equation of the line is:  x+bay+ca=0 

Hence, two parameters should be known.

Correct Option: B


37. The point  (4,1)  undergoes the following two successive transformations:

(i) Reflection about the line  y=x 

(ii) Translation through a distance  2  units along the positive  x axis. Then the final coordinates of the point are

(A)  (4,3) 

(B)  (3,4) 

(C)  (1,4) 

(D)  (72,72) 

Ans: 

Given: (4,1)  undergoes two successive transformations.

The reflection of  A(4,1)  in  y=x is  B(1,4) .

Then, translation of  B through a distance  2  units along the positive  x axis shifts  B  to  C(1+2,4)  i.e.  C(3,4) .

Hence, the final coordinates of point is  (3,4) .

Correct Option: B 


38. A point equidistant from the lines  4x+3y+10=0,5x12y+26=0  and  7x+24y50=0  is

(A)  (1,1) 

(B)  (1,1) 

(C)  (0,0) 

(D)  (0,1) 

Ans: Given: 4x+3y+10=0, 5x12y+26=0 and  7x+24y50=0 .

The distance of  (0,0)  from  4x+3y+10=0 is  |0+0+10|42+32=2 .

The distance of  (0,0)  from  5x12y+26=0 is  |0+0+26|52+122=2 .

The distance of  (0,0)  from  7x+24y50=0 is  |0+050|72+242=2 .

The distance of all the three lines from  (0,0)  is  2 units.

Hence,  (0,0)  is the required point.

Correct Option: C


39. A line passes through  (2,2)  and is perpendicular to the line  3x+y=3 . Its  y intercept is

(A)  13 

(B)  23 

(C)  1 

(D)  43 

Ans:

Given: Line passes through  (2,2) .

The slope of  3x+y=3 is  3 .

So, slope of a line perpendicular to it =13 .

The equation of the required line is:  y2=13(x2)  i.e.  x3y+4=0 .

Put  x=0  to find  y intercept.

 03y+4=0 

 y=43 which is the y intercept.

Hence, the y intercept is  43 .

Correct Option: D


40. The ratio in which the line  3x+4y+2=0  divides the distance between the lines  3x+4y+5=0  and  3x+4y5=0  is

(A)  1:2 

(B)  3:7 

(C)  2:3 

(D)  2:5 

Ans: Given: 3x+4y+5=0 and  3x+4y5=0 .

Distance between  3x+4y+5=0  and  3x+4y+2=0  is  |52|9+16=35 

Distance between  3x+4y5=0  and  3x+4y+2=0  is  |52|9+16=75 

Hence, the ratio is  35:75  i.e.  3:7 .

Correct Option: B


41. One vertex of the equilateral triangle with centroid at the origin and one side as  x+y2=0  is

(A)  (1,1) 

(B)  (2,2) 

(C)  (2,2) 

(D)  (2,2) 

Ans: Given: x+y2=0 

Let  ABC  be the equilateral triangle with vertex  A(h,k) .

The centroid is  G(0,0) .

Slope of line  BC  i.e.  x+y2=0 is  1 .

As AGBC ,

The slope of  AG  is  kh=1 or  h=k .

The distance of  (0,0) from  BC=|0+02|12+12=2 .

Distance of  A  from  BC=32=|h+k2|12+12 

 |h+k2|=6 

 h+k8=0 or h+k+4=0 

 h+h8=0 or h+h+4=0 

 h=4 or h=2 

Hence, the vertex is  (2,2) 

Correct Option:C


42. If  a,b,c  are in A.P., then the straight line  ax+by+c=0  will always pass through.

Ans: Given: a, b, c  are in A.P.

 a, b, c are in A.P.

So,  b=a+c2  i.e.  a2b+c=0 

Compare  a2b+c=0  and line ax+by+c=0 ,

 x=1,y=2 

Hence, the point  (1,2)  lies on the line.


43. The line which cuts off equal intercept from the axes and pass through the point  (1,2)  is

Ans: Given: (1,2) 

The line which cuts equal intercepts from the axes is  xa+ya=1 .

As this line passes through  (1,2), 

 12=a or  a=1 .

Thus, the required equation of line is:  x+y=1 

 x+y+1=0 

Hence, the equation is  x+y+1=0 .


44. Equations of the lines through the point  (3,2)  and making an angle of  45  with the line  x2y=3  are

Ans: Given: Line passes through  (3,2) .

Slope of  x2y=3 is  12

Let the slope of the required line be  m .

 tan45=|m121+12m| 

 1=±2m12+m 

 2m1=2+m or 12m=2+m 

 m=3 or m=13 

The line also passes through (3,2) .

The equation of the line is:  y2=3(x3)  or  y2=13(x3) 

Hence, equations of lines are:  3xy7=0, x+3y9=0 .


45. The points  (3,4)  and  (2,6)  are situated on the  _______  of the line  3x4y8=0 .

Ans: Given: (3,4) and  (2,6) .

Given line is  3x4y8=0 .

For the point  (3,4):  3(3)4(4)8=15<0 

For the point  (2,6):  3(2)4(6)8=22>0 

Hence,  (3,4)  and  (2,6)  lie on opposite side of  3x4y8=0 .


46. A point moves so that square of its distance from the point  (3,2)  is numerically equal to its distance from the line  5x12y=3 . The equation of its locus is 

Ans: Given: (3,2) .

Let the moving point be  P(h,k)

Given point is  A(3,2) .

 AP2=(h3)2+(k+2)2=d12 

The distance of  (h,k)  from  5x12y3=0  is:

 d2=|5h12k325+144| 

 d2=|5h12k313| 

Given that,  d12=d2 

 (h3)2+(k+2)2=5h12k313 

 h26h+9+k2+4k+4=5h12k313 

 13h2+13k278h+52k+169=5h12k3 

 13h2+13k283h+64k+172=0 

Hence, locus is:  13x2+13y283x+64y+172=0 


47. Locus of the mid-points of the portion of the line  xsinθ+ycosθ=p  intercepted between the axes is

Ans: Given: xsinθ+ycosθ=p .

 xsinθ+ycosθ=p meets axes at points A(psinθ,0)  and  B(0,pcosθ) .

Let  P(h,k)  be the mid-point of  AB. 

 h=p2sinθ and k=p2cosθ 

 sinθ=p2h and cosθ=p2k 

By squaring and adding,

 sin2θ+cos2θ=p24h2+p24k2 

 1=p24x2+p24y2 

 4x2y2=p2(x2+y2) 

Hence, the locus is  4x2y2=p2(x2+y2) .


State whether the Statements in Exercises  48  to  56  True or False? Justify.

48. If the vertices of a triangle have integral coordinates, then the triangle cannot be equilateral.

Ans:The given statement is true.

Let us assume an equilateral triangle with vertices  A(x1, y1) ,  B(x2, y2)  and  C(x3, y3) . All the angles of an equilateral triangle are  60 . Assuming the slope of line  AB  as  m1  and the slope of line  BC as  m2 .

 tan60=|m1m21+m1m2| 

 3=|m1m21+m1m2|                       ……  (1) 

Here  m1=y2y1x2x1 and  m2=y3y2x3x2 . Since all the coordinates are integers so  m1  and  m2  are rational numbers.

Therefore, in expression  (1) , the R.H.S is a rational number and the L.H.S is an irrational number. This results in rational number  =  irrational numbers which are not possible.


49. The points  A(2,1)B(0,5)  and  C(1,2)  are collinear.

Ans: The given statement is false.

Using the distance formula, the length  AB 

=(0(2))2+(51)2 

=4+16 

=25 units  

Similarly, the length  BC 

=(10)2+(25)2 

=1+9 

=10 units  

And, the length  AC 

=(2(1))2+(12)2 

=1+1 

=2 units  

Clearly the sum of any two line segments is not equal to the third one in any case, so the points are not collinear.


50. Equation of the line passing through the point  (acos3θ,asin3θ)  and perpendicular to the line  xsecθ+ycosecθ=a is  xcosθysinθ=asin2θ .

Ans:The given statement is false.

The slope of the given line  xsecθ+ycosecθ=a  is,

m1=secθcosecθ 

m1=(1cosθ)(1sinθ) 

m1=(sinθcosθ)  

Therefore, the slope of the required line which is perpendicular to the given line is,

m2=1m1 

m2=1(sinθcosθ)  

m2=cosθsinθ  

So the equation of the line passing through the point  (acos3θ, asin3θ)  and having slope  m2  is,

(yasin3θ)=m2(xacos3θ) 

(yasin3θ)=cosθsinθ(xacos3θ) 

sinθ(yasin3θ)=cosθ(xacos3θ) 

xcosθysinθ=acos4θasin4θ  

Using the algebraic identity  m2n2=(m+n)(mn)  and the trigonometric identity  sin2θ+cos2θ=1 ,

xcosθysinθ=a(cos2θsin2θ)(cos2θ+sin2θ) 

xcosθysinθ=a(cos2θsin2θ)  

Using the identity  cos2θsin2θ=cos2θ ,

xcosθysinθ=acos2θ 


51. The straight line  5x+4y=0  passes through the point of intersection of the straight lines  x+2y10=0 and  2x+y+5=0 .

Ans: The given statement is true.

The intersection of the lines  x+2y10=0  and  2x+y+5=0  is their solution coordinates. Solving the two equations algebraically gives,

x=(203) and  y=(253) , so the solution coordinate is  (203, 253) .

Now, if the line  5x+4y=0  passes through the above coordinates then it must satisfy the point. On substitution,

5×(203)+4×(253)=0 

(1003)+(1003)=0 

0=0  

Hence, the point satisfies the line.


52. The vertex of an equilateral triangle is  (2,3)  and the equation of the opposite side is  x+y=2 . Then the other two sides are  y3=(2±3)(x2) .

Ans: The given statement is true.

Let us assume the slope of one of the remaining two sides as  m . The slope of the given line  x+y=2 is  1 . If  m1  and  m2  are slopes of two line segments then the angle  (θ)  between them is given as  tanθ=|m1m21+m1m2| .

The angle between any two sides of an equilateral triangle is  60 ,

tan60=|1m1+(1)×m| 

3=|(1+m)1m| 

3=|1+m1m|  

Removing the modulus sign,

 ± 3=(1+m1m) 

  1. Considering the positive sign,

3=(1+m1m)

m=(313+1)

m=(23) (By rationalization)

  1. Considering the negative sign,

(3)=(1+m1m) 

m=(3+131)  

m=(2+3) (By rationalization)

Hence, there are two values of  m  for the two remaining sides of the triangle. Both the sides will pass through the point  (2, 3) . So, equation of the other two sides are  (y3)=(2±3)(x2) .


53. The equation of the line joining the point  (3,5)  to the point of intersection of the lines  4x+y1=0  and  7x3y35=0  is equidistant from the points  (0,0) and  (8,34) .

Ans: The given statement is true.

Solving the equations  4x+y1=0  and  7x3y35=0  algebraically gives  x=2 and  y=7 . So the point of intersection is  (2, 7) .

The equation of the line passing through the points (3, 5) and  (2, 7)  is,

(y5)=(7523)(x3) 

(y5)=(12)(x3) 

12xy31=0  

Now, the distance of the point  (0, 0)  from the line  12xy31=0 ,

=|(0×12)+(0×(1))31122+(1)2| 

=31145  

Similarly, the distance of the point  (8, 34)  from the line  12xy31=0 ,

=|(8×12)+(34×(1))31122+(1)2| 

=31145  

Hence, the points are equidistant from the required line.


54. The lines  xa+yb=1  moves in such a ways that  1a2+1b2=1c2 , where  c  is a constant. The locus of the foot of the perpendicular from the origin on the given line is  x2+y2=c2 .

Ans: The given statement is true.

Let the coordinates of the foot of perpendicular drawn from the origin (0,0) is (x,y).

The given line is xa+yb=1 which can be written as xa+yb1=0.

The foot of perpendicular is given as,

(x0)(1a)=(y=0)(1b)=(1a×0)+(1b×0)1(1a)2+(1b)2

ax=by=1(1a2+1b2)

Using the given relation 1a2+1b2=1c2,

ax=by=1(1c2)

ax=by=(c2)

Equating the relations ax=(c2) and by=(c2) gives

x=(c2a)

and y=(c2b), respectively.

On squaring and adding,

x2+y2=(c4a2)+(c4b2)

x2+y2=c4(1a2+1b2)

x2+y2=c4(1c2)

x2+y2=c2


55. The lines ax + 2y + 1 = 0 , bx + 3y + 1 = 0  and cx + 4y + 1 = 0  are concurrent if a, b  and c  are in G.P.

Ans: The given statement is false.

The given lines can be written are ax+2y+1=0, bx+3y+1=0 and cx+4y+1=0. For these to be concurrent it must satisfy the determinant relation,

$\Rightarrow\left|\begin{array}{lll}

a & 2 & 1 \\

b & 3 & 1 \\

c & 4 & 1

\end{array}\right|=0$

Performing the row operations R2R2R1 and R3R3R1,

$\Rightarrow\left|\begin{array}{ccc}

a & 2 & 1 \\

b-a & 1 & 0 \\

c-a & 2 & 0

\end{array}\right|=0$

Expanding the determinant along C3,

2(ba)(ca)=0

2bac=0

a+c=2b

Therefore, for the lines to be concurrent a,b and c must be in A.P.


56. Line joining the points  (3,4)  and  (2,6)  is perpendicular to the line joining the points  (3,6) and  (9,18) .

Ans: The given statement is false.

The line joining the points  (3, 4)  and  (2, 6)  is,

(y6)=(6(4)23)(x(2)) 

(y6)=(2)(x+2) 

y=2x+2   (9, 18) 

The slope of this line is  2 .

Similarly, the line joining the points  (3, 6)  and  is,

(y6)=(6(18)39)(x(3)) 

(y6)=(2)(x+3) 

y=2x  

The slope of this line is also  2 .

For the two lines to be perpendicular the product of their slopes must be  1 , but here the product is  (2)×(2)=4 .


Match the questions given under Column  C1  with their appropriate answers given under the column  C2  in Exercises  57 to  59 .

57.

Column  C1 

Column C2 

(a) The coordinates of the points

 P  and  Q  on the line  x+5y=13  which

are at a distance of  2  units from the line

 12x5y+26=0  are

(i)  (3,1)(7,11) 

(b) The coordinates of the points on the line 

 x+y=4  which are at a unit distance

from the line  4x+3y10=0  are

(ii)  (13,113)(43,73) 

(c) The coordinates of the points on the line

joining  A(2,5)  and  B(3,1)  such

that  AP=PQ=QB  are

(iii)  (1,125)(3,165) 


Ans:

(a) Let one of the required point is  (a, b) . Since it lies on the line  x+5y=13 , so it will satisfy the line.

 a+5b=13                       ……  (1) 

The distance of the point  (a, b)  from the line  12x5y+26=0  is  2  units,

|(12×a)+((5)×b)+26(12)2+(5)2|=2 

|12a5b+26|=26 

(12a5b+26)=± 26  

Considering the positive sign and solving equation  (1)  and  (12a5b+26)=26  gives the point  (a, b)=(1, 125) .

Considering the negative sign and solving equation  (1)  and  (12a5b+26)=26  gives the point  (a, b)=(3, 165) .


(b) Let one of the required point is  (c, d) . Since it lies on the line  x+y=4 , so it will satisfy the line.

 c+d=4                       ……  (1) 

The distance of the point  (c, d)  from the line  4x+3y10=0  is  1  unit,

|(4×c)+(3×d)10(4)2+(3)2|=1 

|4c+3d10|=5 

(4c+3d10)=± 5  

Considering the positive sign and solving equation  (1)  and  (4c+3d10)=5  gives the point  (c, d)=(3, 1) .

Considering the negative sign and solving equation  (1)  and  (4c+3d10)=5  gives the point  (c, d)=(7, 11) .


(c) Here the points  P  and  Q  trisect the line  AB . Let the points be  P(x1, y1) and  Q(x2, y2) .

For point  P ,  AP:PB=m:n=1:2 . The section formula states that if a point  (x,y)  divides a line segments joining the two points  (x1,y2)  and  (x2,y2)  in the ratio  m:n  then,

(x, y)=(mx2+nx1m+n, my2+ny1m+n) 

Using this formula for the points  A ,  P  and  B ,

(x1, y1)=((3×1)+(2×(2))1+2, (1×1)+(5×2)1+2) 

(x1, y1)=(13, 113)  

Now,  P  is the midpoint of  AQ . The midpoint formula is a special case of the section formula with  m:n=1:1 .

Using the midpoint formula for  A ,  P  and  Q , (13, 113)=(2+x22, 5+y22) 

Solving for the coordinates of  Q  by comparison gives  (x2, y2)=(43, 73) .

Hence, the correct match is (a) – (iii), (b) – (i), (c) – (ii).


58. The value of the  λ , if the lines  (2x+3y+4)+λ(6xy+12)=0  are


Column  C1 

Column C2 

(a) Parallel to  y  axis is

(i)  λ=34 

(b) Perpendicular to  7x+y4=0  are

(ii)  λ=13 

(c) Passes through  (1,2)  is

(iii)  λ=1741 

(d) Parallel to  x  axis

(iv)  λ=3 


Ans: Simplifying the line  (2x+3y+4)+λ(6xy+12)=0 

gives  (2+6λ)x+(3λ)y+(4+12λ)=0 .

(a) For the line  (2+6λ)x+(3λ)y+(4+12λ)=0  to be parallel to the  y  axis, the coefficient of  y  coordinate must be  0 .

(3λ)=0 

λ=3  


(b) The slope of the line  7x+y4=0 is  7 , therefore the slope of the line  (2+6λ)x+(3λ)y+(4+12λ)=0  must be  17  so that the two lines are perpendicular.

(2+6λ)(3λ)=17 

1442λ=3λ 

λ=1741  


(c) If the point  (1, 2)  passes through the line  (2+6λ)x+(3λ)y+(4+12λ)=0  then it must satisfy the equation. On substitution,

(2+6λ)+(62λ)+(4+12λ)=0 

16λ=12 

λ=34  


(d) For the line  (2+6λ)x+(3λ)y+(4+12λ)=0  to be parallel to the  x  axis, the coefficient of  x  coordinate must be  0 .

(2+6λ)=0 

λ=13  

Hence, the correct match is (a) – (iv), (b) – (iii), (c) – (i), (d) – (ii).


59. The equation of the line through the intersection of the lines  2x3y=0  and  4x5y=2  and

Column C1 

Column C2 

(a) Through the point  (2,1)  is

(i)  2xy=4 

(b) Perpendicular to the line

 x+2y+1=0  is

(ii)  x+y5=0 

(c) Parallel to the line

 3x4y+5=0  is

(iii)  xy1=0 

(d) Equally inclined to the axes is

(iv)  3x4y1=0 


Ans: The point of intersection of the lines  2x3y=0  and  4x5y=2  is obtained by solving the equations algebraically. So the intersection point is  (3, 2) .

(a) The equation of the line passing through the points  (2, 1)  and  (3, 2)  is,

(y1)=(2132)(x2) 

(y1)=(x2) 

xy1=0  


(b) The slope of the line  x+2y+1=0 is  (12) , so the slope of the line perpendicular to this line will be  2 . Now, the line passes through the point  (3, 2)  so its equation is,

(y2)=2(x3) 

(y2)=(2x6) 

2xy=4  


(c) The slope of the line  3x4y+5=0 is  (34) , so the slope of the line parallel to this line will also be  (34) . Now, the line passes through the point  (3, 2)  so its equation is,

(y2)=(34)(x3) 

(4y8)=(3x9) 

3x4y1=0  


(d) The lines which are equally inclined to the axes have their slope values equal to  tan45=1 or  tan135=1 . Now, , the line passes through the point  (3, 2)  so its equation is,

Considering the positive slope,

(y2)=(1)(x3) 

xy1=0  

Considering the negative slope,

(y2)=(1)(x3) 

x+y5=0  

Hence, the correct match is (a) – (iii), (b) – (i), (c) – (iv), (d) – (ii) or (iii).


About the Chapter

A straight line is defined as a combination of an infinite number of points joined at both ends of a point. In other words,  it is an endless one-dimensional figure that does not have any curve but the ends can extend up to infinity. A straight line can be horizontal, vertical and slanted and have equations associated with the same. Any straight line, on a plane, will always be measured at 180° for an angle drawn between two points on the same straight line. A straight line can also be called the shortest distance between two points and is represented on the graph as A(x1, y1) and B(x2, y2). There are three kinds of straight lines:

  1. A Horizontal Line is parallel to the x-axis and perpendicular to the y axis. The angle formed is 0° or 180° with the x-axis and 90° or 270° with the y axis on the graph.

  2. A Vertical Line is parallel to the y axis and perpendicular to the x-axis. The angle formed is 90° or 270° with the x-axis and 0° or 180° with the y axis on the graph.

  3. A Slanted Line can form any angle other than 0°, 90°, 180°, 270° and 360° with the horizontal and vertical lines on the graph. 

Other details like equations of the slopes formed by the straight lines, equations of lines parallel to the x-axis or y-axis, various forms of the equation of a line, etc., have also been dealt with.


NCERT Exemplar for Class 11 Maths Chapter 10 - Straight Lines is divided into different sections from 10.1 to 10.1.8. Section 10.2 deals with examples on the different topics of straight lines and Section 10.3 comprises exercises/question bank.

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FAQs on NCERT Exemplar for Class 11 Maths Chapter 10 - Straight Lines (Book Solutions)

1. What is the content of NCERT Exemplar for Class 11 Maths Chapter 10 - Straight Lines?

Chapter 10 Straight Lines in NCERT class 11 Maths has been explained conceptually so that the lesson is etched in the student’s memory without fail. Each topic has been concisely analysed and presented with examples and different methods to solve different questions. This chapter is split into several segments:

  • 10.1- Overview of Straight Lines

  • 10.1.1- Slope of a Line

  • 10.1.2- Angle between two lines

  • 10.1.3- Collinearity of three points

  • 10.1.4- Various forms of the equation of a line

  • 10.1.5- General equation of a line

  • 10.1.6- Distance of a point from a line

  • 10.1.7- Locus and Equation of a Locus

  • 10.1.8- Intersection of two given lines

2. What are the various forms of the equation of a line?

The general equation of a straight line is ax+by+c=0 where a, b and c are constants, x and y are variables and the slope is -a/b. The various slopes associated with this equation are as follows:

  • Slope and Y-Intercept Form where slope m = tanθ. θ is the angle formed by the line with the positive x-axis, and y-intercept as b is given by: y = mx + b, where m is the slope.

  • Slope Point Form in which a straight line passes through a point where y-y1= m(x-x1)

  • Two Points Form in which a straight line passed through two points where (y-y1)= [(y2-y1)/x2-x1](x-x1)

  • Intercept Form in which one point A is on the y axis and point B is on the x-axis where (x/a)+(y/b)=1

3. What are the different kinds of slopes of a straight line?

There are four different kinds of slopes formed by a straight line. They are as follows:

  • Zero slopes in which the angle formed by the straight line is 0° with the x-axis, ie., it is parallel to the x-axis.

  • Positive Slope in which the angle formed with its x axis is between 0° and 90° hence the slope of the line is positive.

  • Negative Slope in which the angle formed with its x-axis is between 90° and 180° hence the slope of the line is negative

  • Infinite Slope in which the line forms a 90° angle with the x-axis or lies parallel to the y axis. Then the slope is not defined or is infinite.

4. Where can we download NCERT Exemplar for Class 11 Maths Chapter 10 - Straight Lines?

Since NCERT books are available online for free, students can easily access the book at ncert.nic.in and download the book in PDF format. Vedantu has also created study materials on NCERT Exemplar for Class 11 Maths Chapter 10 - Straight Lines (Book Solutions) with solved questions and examples for students to understand the concepts even better. Instead of searching on the internet for hours for resources and other study materials, students can click on Vedantu's official website and download the same for free. 

5. Can solving the Previous Years’ Question Paper clear all doubts on Straight Lines?

Previous Year’s Question Papers are very important to solve since they help a student boost his/her confidence and also solving different levels of questions helps them enhance their performance. They are an important tool for revision that students can practice after solving the exercises in the NCERT textbook. It also helps the student keep a track of their strengths and weaknesses for improvement and provides a set pattern of the exam and the weightage of marks on each question.