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NCERT Solutions for Class 11 Maths Chapter 10: Straight Lines - Exercise 10.3

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NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines

Free PDF download of NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.3 (Ex 10.3) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 11 Maths Chapter 10 Straight Lines Exercise 10.3 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails.


Class:

NCERT Solutions for Class 11

Subject:

Class 11 Maths

Chapter Name:

Chapter 10 - Straight Lines

Exercise:

Exercise - 10.3

Content-Type:

Text, Videos, Images and PDF Format

Academic Year:

2023-24

Medium:

English and Hindi

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Access NCERT Solutions for Class 11 Mathematics Chapter 10-Straight Lines

Exercise 10.3

1. Reduce the following equation into slope-intercept form and find their slopes and the $y$ intercepts.

i. $x  +  7y  =  0$ 

Ans: The equation is $x+7y=0$.

We can write it as $y  =  -  \dfrac{1}{7}  x  +  0$$\ldots (1)$

This equation is of the form $y=mx+c$, where $m=-\dfrac{1}{7}$ and $c=0$ 

Therefore, the equation $\left( 1 \right)$ is the slope-intercept form, where the slope and the y-intercept are $-\dfrac{1}{7}$ and $0$ respectively.

ii. $6x+3y-5=0$ 

Ans: The given equation is $6x+3y-5=0$. 

We can write it as $y=\dfrac{1}{3}(-6x+5)$

$\Rightarrow y=-2x+\dfrac{5}{3}\quad \ldots (1)$

This equation is of the form$\text{y}=\text{mx}+\text{c}$, where $\text{m}=-2$ and$\text{c}=\dfrac{5}{3}$. Therefore, equation $\left( 1 \right)$  in the slope-intercept form, where the slope and the $y$-intercept are $-2$ and $\dfrac{5}{2}$ respectively.

iii. $y=0$

Ans: The given equation is $y=0$. 

We can write it as $y=0.x+0$$.....\left( 1 \right)$

This equation is of the form$y=mx+c$, where $m=0$ and $c=0$. Therefore, equation $\left( 1 \right)$ is in the slope-intercept form, where the slope and the $y$-intercept are \[0\] and \[0\] respectively.


2. Reduce the following equations into intercept form and find their intercepts on the axes.

i. $3x+2y-12=0$

Ans: The given equation is $3x+2y-12=0$.

We can write it as,

$3x+2y=12$

$\Rightarrow \dfrac{3x}{12}+\dfrac{2y}{12}=1$

i.e. $\dfrac{x}{4}+\dfrac{y}{6}-1\quad \ldots (1)$

This equation is of the form$\dfrac{x}{a}+\dfrac{y}{b}=1$, where $a=4$ and$b=6$. 

Therefore, equation (1) is in the intercept form, where the intercepts on the $\text{x}$ and $\text{y}$ axes are $4$ and $6$ respectively.

ii. $4x-3y=6$

Ans:  The given equation is $4x-3y=6$. 

We can write it as,

$\dfrac{4x}{6}-\dfrac{3y}{6}=1$

$\Rightarrow \dfrac{2x}{3}-\dfrac{y}{2}=1$

i.e. $\dfrac{x}{\left( \dfrac{3}{2} \right)}+\dfrac{y}{(-2)}=1$$\ldots (2)$

Therefore, equation $(2)$ is in the intercept form, where the intercepts on $\text{x}$ and $\text{y}$ axes are $\dfrac{3}{2}$ and $-2$ respectively.

iii. $3y+2=0$.

Ans: The given equation is $3\text{y}+2=0$. 

We can write it as $3y=-2$

i.e. $\dfrac{\text{y}}{\left( -\dfrac{2}{3} \right)}=1$$\ldots (3)$

Therefore, equation is in the$\dfrac{x}{a}+\dfrac{y}{b}=1$, where $a=0$ and$b=-\dfrac{2}{3}$. 

Therefore, equation (3) is in the intercept form, where the intercept on the $y$-axis is $-\dfrac{2}{3}$ .

 It has no intercept on the $x$-axis.


3. Reduce the following equations into normal form. Find their perpendicular distance from the origin and angle between perpendicular and the positive $x$-axis.

i. $x-\sqrt{3}y+8=0$

Ans: The given equation is $x-\sqrt{3}y=-8$

$\Rightarrow -x+\sqrt{3}y=8$

By dividing both sides with$\sqrt{{{(-1)}^{2}}+{{(\sqrt{3})}^{2}}}=\sqrt{4}=2$, we obtain $-\dfrac{x}{2}+\dfrac{\sqrt{3}}{2}y=\dfrac{8}{2}$

$\Rightarrow \left( -\dfrac{1}{2} \right)x+\left( \dfrac{\sqrt{3}}{2} \right)y=4$

$\Rightarrow x\cos {{120}^{2}}+y\sin {{120}^{{}^\circ }}=4\quad .....(1)$

The Equation $\left( 1 \right)$is in the normal form.

Compare the equation$\left( 1 \right)$with the normal form of the equation of the line$x\cos \omega   +  y\sin \omega   =  p$, we get

 $\omega ={{120}^{{}^\circ }}$ and$p=4$.

Therefore, the perpendicular distance of the line from the origin is $4$ .

Moreover, the angle between the perpendicular and the positive $x-$axis is$120{}^\circ $.

iii. $y-2=0$

Ans: The given equation is $y-2-a$.

We reduce the equation as  $0\text{x}+\text{t}\cdot \text{y}=2$

By dividing both sides with$\sqrt{{{0}^{2}}+{{1}^{2}}}=1$, we get

 $0.x+1-y=2$ 

$\Rightarrow x\cos 90{}^\circ +y\sin 90{}^\circ =2$$\ldots (2)$

The Equation $\left( 2 \right)$ is in the normal form. 

Compare the equation (2) with the normal form of the equation of the line$x\cos \omega +y\sin \omega =p$,

 We get,

 $\omega ={{90}^{{}^\circ }}$ and$p=2$.

Therefore, the perpendicular distance of the line from the origin is$2$.

Moreover, the angle between the perpendicular and the positive $x$-axis is ${{90}^{{}^\circ }}$ .

iv. $x-y=4$

Ans: The given equation is$x-y=4$.

We reduce the equation as $1x+(-1)y=4$

Divide both sides with$\sqrt{{{1}^{2}}+{{(-1)}^{2}}}=\sqrt{2}$, 

We get,

$\dfrac{1}{\sqrt{2}}x+\left( -\dfrac{1}{\sqrt{2}} \right)y=\dfrac{4}{\sqrt{2}}$

$\Rightarrow x\cos \left( 2\pi -\dfrac{\pi }{4} \right)+y\sin \left( 2\pi -\dfrac{\pi }{4} \right)=2\sqrt{2}$

$\Rightarrow x\cos 315{}^\circ +y\sin 315{}^\circ =2\sqrt{2}.....(3)$

The Equation $(3)$ is in the normal form.

Compare the equation $(3)$with the normal form of the equation of the line$x\cos \omega +y\sin \omega =p$, 

We get $\omega =315$  and$\text{p}=2\sqrt{2}$.

Thus, the perpendicular distance of the line from the origin is$2\sqrt{2}$, while the angle between the perpendicular and the positive $x$-axis is$315{}^\circ $.


4. Find the distance of the points $(-1,1)$ from the line $12(x+6)=5(y-2)$.

Ans: The equation of the line is$12(x+6)=5(y-2)$. 

$\Rightarrow 12x+72=5y-10$

$\Rightarrow 12x-5y+8z=0$

Compare the equation $\left( 1 \right)$ with  the general equation of the line $Ax+By+C=0$, we obtain $A=12$, $\text{B}=-5$, and $C=82$.

The perpendicular distance (d) of a line $Ax+By+C=0$ from a point $\left( {{x}_{1}},{{y}_{1}} \right)$ is  

\[\text{d}=\dfrac{\left| \text{A}{{\text{x}}_{1}}+B{{y}_{1}}+\text{C} \right|}{\sqrt{{{A}^{2}}+{{B}^{2}}}}\]

The given point is$\left( {{x}_{1}},{{y}_{1}} \right)=(-1,1)$. 

So, the distance of the point $(-1,1)$ from the given line

$=\frac{|12(-1)+(-5)(1)+82|}{\sqrt{(12)^{2}+(-5)^{2}}}$ units $=\frac{|-12-5+82|}{\sqrt{169}}$ units $=\frac{|65|}{13}$ units $=5$ units


5. Find the points on the x-axis at what distance from the line $\dfrac{x}{3}+\dfrac{y}{4}=1$ is $4$ units.

Ans: The given equation of the line is $\dfrac{x}{3}+\dfrac{y}{4}=1$

Or $4x+3y-12=0$

Compare the equation (1) with the general equation of the line$Ax+By+C=0$.

We get,

$A=4$,$B=3$, and $C=-12$

Let $(a,0)$ be the point on the x-axis whose distance from the given line is $4$units.

The perpendicular distance $(d)$ of a line $Ax+By+C=0$ from a point $\left( {{x}_{1}},{{y}_{1}} \right)$ is $d=\dfrac{\left| A{{x}_{1}}+B{{y}_{1}}+c \right|}{\sqrt{{{A}^{2}}+{{B}^{2}}}}$

Then  $4=\dfrac{\left| 4a+3\times 0-12 \right|}{\sqrt{{{4}^{2}}+{{3}^{2}}}}$

$\Rightarrow \text{  }4=\dfrac{\left| 4a-12 \right|}{5}$

$\Rightarrow \text{  }|4a-12|=20$

$\Rightarrow \text{  }\pm (4a-12)=20$

$\Rightarrow (4a-12)=20$ or $+(4a-12)=20$

$\Rightarrow 4a=20+12$ or $4{{a}^{-}}-20+12$

$\Rightarrow a=8$ or $-2$

Thus, the required points on $x$the -axis are $(-2,0)$ and $(8,0)$.


6. Find the distance between parallel lines

i. $15x+8y-34=0$ and $15x+8y+31=0$

Ans: The distance $\left( d \right)$between parallel lines $Ax  +  By  +{{C}_{1}}  =  0$ and $\text{Ax}+\text{By}+{{\text{C}}_{2}}=0$ is$d=\dfrac{\left| {{C}_{1}}-{{C}_{2}} \right|}{\sqrt{{{A}^{2}}+{{B}^{2}}}}$.

The parallel lines are $15x+8y-34=0$ and $15x+8y+31=0$ .

Here, 

$A=15,~\text{B}=8,{{C}_{1}}=-34$, and${{C}_{2}}=31$.

Therefore, the distance between the parallel lines is

\[d  =  \dfrac{\left| {{C}_{1}}  -  {{C}_{2}} \right|}{\sqrt{{{A}^{2}}  +  {{B}^{2}}}}  =  \dfrac{\left| -34  -31 \right|}{\sqrt{{{15}^{2}}  +  {{8}^{2}}}}  \]

\[=  \dfrac{\left| -65 \right|}{\sqrt{289}}  \]Units.

\[=  \dfrac{65}{17}  \]Units.

ii. $l(x+y)+p=0$ and $l(x+y)-r-0$

Ans: The distance $\left( d \right)$between parallel lines $Ax  +  By  +{{C}_{1}}  =  0$ and $\text{Ax}+\text{By}+{{\text{C}}_{2}}=0$ is given by,

$d=\dfrac{\left| {{C}_{1}}-{{C}_{2}} \right|}{\sqrt{{{A}^{2}}+{{B}^{2}}}}$.

The parallel lines are $l(\text{x}+y)+\text{p}=0$ and $l(\text{x}+\text{y})-\text{r}=0$

i.e $lx+ly+p=0$ and $lx+ly-r=0$

Here,

$\text{A}=l ,\text{B}=l,{{C}_{1}}=p$, and ${{C}_{2}}=-{r}$

Therefore, the distance between the parallel lines is

\[d  =  \dfrac{\left| {{C}_{1}}  -  {{C}_{2}} \right|}{\sqrt{{{A}^{2}}  +  {{B}^{2}}}}  =  \dfrac{\left| p  +  r \right|}{\sqrt{{{l}^{2}}  +  {{l}^{2}}}}  \]\[=  \dfrac{\left| p  +  r \right|}{\sqrt{2 {{l}^{2}}}}  \]Units.

\[=  \dfrac{\left| p  +  r \right|}{l\sqrt{ 2}}  \]Units.

\[\Rightarrow d  =  \dfrac{1}{\sqrt{2}}  \dfrac{\left| p  +  r \right|}{l}  \]Units.


7. Find the equation of the line parallel to the line $3x-4y+z=0$ and passing through the point$\left( -  2,3 \right)$.

Ans: The equation of the line is given as,

$3x-4y+2=0$

Or $y  =  \dfrac{3x}{4}  +  \dfrac{2}{4}$

or $y=\dfrac{3}{4}x+\dfrac{1}{2}$ 

Which is of the form $y=nx+0$

$\therefore $ The slope of the given line $=\dfrac{3}{4}$

 It is known that parallel lines have the same slope.

$\therefore $ The slope of the other line $=m=\dfrac{3}{4}$ 

Now, the equation of the line that has a slope of $\dfrac{3}{4}$ and passes through the points $(-2,3)$ is

$(y-3)=\dfrac{3}{4}\{x-(-2)\}$

$\Rightarrow 4y-12=3x+6$

i.e ,$3x-4y+18=0$


8. Find the equation of the line perpendicular to the line $x-7y+5=0$ and having $x$ intercept $3$.

Ans: The equation of the line is $x-7y+5=0$.

Or $y=\dfrac{1}{7}x+\dfrac{5}{7}$,

 which is of the form $y=mx+c$

$\therefore $ The slope of the given line $=\dfrac{1}{7}$ .

The slope of the line perpendicular to the line having a slope of $\dfrac{1}{7}$ is $m=-\dfrac{1}{\left( \dfrac{1}{7} \right)}=-7$.

 The equation of the line with slope $-7$ and $x$-intercept 3 is given by $y=m(x-d)$

$\Rightarrow y=-7(x-3)$

$\Rightarrow y=-7x+21$

$\Rightarrow 7x+y=21$


9. Find angles between the lines $\sqrt{3}x+y=1$ and $x+\sqrt{3}y=1$.

Ans: The given lines are $\sqrt{3}x+y=1$ and $x+\sqrt{3}y=1$

 $y=-\sqrt{3x}+1 \quad -(1)$  and $y=-\dfrac{1}{\sqrt{3}}x+\dfrac{1}{\sqrt{3}}\quad -(2)$

The slope of the line (1) is ${{m}_{1}}=-\sqrt{3}$, while the slope of the line (2) is ${{m}_{2}}=-\dfrac{1}{\sqrt{3}}$.

 The acute angle  between the two lines is given by

$\tan \theta =\left| \dfrac{{{m}_{1}}-{{m}_{2}}}{1+m,{{m}_{2}}} \right|$

$\tan \theta =\left| \dfrac{-\sqrt{3}+\dfrac{1}{\sqrt{3}}}{1+(-\sqrt{3})\left( -\dfrac{1}{\sqrt{3}} \right)} \right|$

$\tan \theta =\left| \dfrac{\dfrac{-3+1}{\sqrt{3}}}{1+1} \right|=\left| \dfrac{-2}{2\times \sqrt{3}\mid } \right|$

$\tan \theta =\dfrac{1}{\sqrt{3}}$

$\theta ={{30}^{{}^\circ }}$

Thus, the angle between the given lines is either ${{30}^{{}^\circ }}$ or ${{180}^{*}}-{{30}^{{}^\circ }}={{150}^{{}^\circ }}$.


10. The line through the points $(\text{h},3)$ and $(4,1)$ intersects the line $7\text{x}-9\text{y}-19-0$. At a right angle. Find the value of $\text{h}$.

Ans: The slope of the line passing through points $(\text{h},3)$ and $(4,1)$ is 

${{m}_{1}}=\dfrac{1-3}{4-h}=\dfrac{-2}{4-h}$

The slope of the line $7\text{x}-9y-19=0$ or $y=\dfrac{7}{9}x-\dfrac{19}{9}$ is ${{\text{m}}_{2}}=\dfrac{7}{9}$.

It is given that the two lines are perpendicular. 

$\therefore \Rightarrow \text{  }{{m}_{1}}\times {{m}_{2}}=-1$

$\Rightarrow \dfrac{-14}{36-9h}=-1$

$\Rightarrow \text{  }14=36-9h$

$\Rightarrow 9h=36-14$

$\Rightarrow h=\dfrac{22}{9}$

Thus, the value  $h$ is $\dfrac{22}{9}$


11. Prove that the line through the point $\left( {{x}_{1}},{{y}_{1}} \right)$ and parallel to the line $Ax+By+C=0$ is. $A\left( x-{{x}_{1}} \right)+B\left( y-{{y}_{1}} \right)=0$

Ans: The slope of the line $Ax+By+C=0$ or $y=\left( \dfrac{-A}{B} \right)x+\left( \dfrac{-C}{B} \right)$ is $m=-\dfrac{A}{B}$ .

The parallel lines have the same slope.

$\therefore $ The slope of the other line $m=-\dfrac{A}{B}$

The equation of the line passing through a point $\left( {{x}_{1}},{{y}_{1}} \right)$ and having a slope $m=-\dfrac{A}{B}$ is $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$

$y-{{y}_{1}}=-\dfrac{A}{B}\left( x-{{x}_{1}} \right)$

$B\left( y-{{y}_{1}} \right)=-A\left( x-{{x}_{1}} \right)$

$A\left( x-{{x}_{1}} \right)+B\left( y-{{y}_{1}} \right)=0$

Hence, the line through point $\left( {{x}_{1}}-{{y}_{1}} \right)$ and parallel to line $\text{Ax}+\text{By}+C=0$ is $A \left( x-{{x}_{1}} \right)+B\left( y-{{y}_{1}} \right)=0$


 12. Two lines passing through the points $(2,3)$ intersect each other at an angle

 of $60{}^\circ $. If the slope of the one line is $2$, find the equation of the other line.

Ans: It is given that the slope of the first line, ${{m}_{1}}=2$. 

Let the slope of the other line be ${{m}_{2}}$ .

The angle between the two lines is ${{60}^{{}^\circ }}$.

$\therefore \tan 60{}^\circ =\left| \dfrac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|$

$\Rightarrow \sqrt{3}=\left| \dfrac{2-{{m}_{2}}}{1+2{{m}_{2}}} \right|$

$\Rightarrow \sqrt{3}=\pm \left( \dfrac{2-{{m}_{2}}}{1+2{{m}_{2}}} \right)$

$\Rightarrow \sqrt{3}=\dfrac{2-{{m}_{2}}}{1+2{{m}_{2}}}  \text{or}  \sqrt{3}=-\left( \dfrac{2-{{m}_{3}}}{1+2{{m}_{2}}} \right)$

$\Rightarrow \sqrt{3}\left( 1+2{{m}_{2}} \right)=2-{{m}_{2}}$ or $\sqrt{3}\left( 1+2{{m}_{2}} \right)=-\left( 2-{{m}_{2}} \right)$

\[\Rightarrow \sqrt{3}+2\sqrt{3}{{m}_{2}}+{{\text{m}}_{2}}=2\] or \[\sqrt{3}+2\sqrt{3}{{m}_{2}}-{{m}_{2}}  =  -2\]

$\Rightarrow \sqrt{3}+(2\sqrt{3}+1){{\text{m}}_{2}}=2$or $\sqrt{3}+(2\sqrt{3}-1){{\text{m}}_{3}}=-2$

$\Rightarrow {{m}_{2}}=\dfrac{2-\sqrt{3}}{(2\sqrt{3}+1)}  \text{or}  {{m}_{2}}=\dfrac{-(2+\sqrt{3})}{(2\sqrt{3}-1)}$

Case 1: $\quad {{\text{m}}_{2}}=\left( \dfrac{2-\sqrt{3}}{(2\sqrt{3}+1)} \right)$

The equation of the line passing through the point $(2,3)$ and having a slope of $\dfrac{2-\sqrt{3}}{(2\sqrt{3}+1)}$is 

 $(y-3)=\dfrac{(2-\sqrt{3})}{(2\sqrt{3}+1)}(x-2)$

$(2\sqrt{3}+1)y-3(2\sqrt{3}+1)=(2-\sqrt{3})x-(2-\sqrt{3})2$

$(\sqrt{3}-2)x+(2\sqrt{3}+1)y=-4+2\sqrt{3}+6\sqrt{3}+3$

$(\sqrt{3}-2)x+(2\sqrt{3}+1)y=-1+8\sqrt{3}$

In this case, the equation of the other line is $(\sqrt{3}-2)x+(2\sqrt{3}+1)y=-1+8\sqrt{3}$.

Case 2: $\quad {{\text{m}}_{2}}=\left( \dfrac{-\left( 2+\sqrt{3} \right)}{(2\sqrt{3}-1)} \right)$

The equation of the line passing through the point $(2,3)$ and having a slope of $\left( \dfrac{-\left( 2+\sqrt{3} \right)}{(2\sqrt{3}-1)} \right)$ is 

$(y-3)=\dfrac{-(2+\sqrt{3})}{(2\sqrt{3}-1)}(x-2)$

$(2\sqrt{3}-1)y-3(2\sqrt{3}-1)=(2\sqrt{3}-1)x+2(2\sqrt{3}-1)$

$(2\sqrt{3}-1)y+(2\sqrt{3}-1)x=4+2\sqrt{3}+6\sqrt{3}-3$

$(2+\sqrt{3})x+(2\sqrt{3}-1)y=1+8\sqrt{3}$

In this case, the equation of the other line is$(2+\sqrt{3})x+(2\sqrt{3}-1)y=1+8\sqrt{3}$

The required equation of the line is  $(\sqrt{3}-2)x+(2\sqrt{3}+1)y=-1+8\sqrt{3}$ or $(2+\sqrt{3})x+(2\sqrt{3}-1)y=1+8\sqrt{3}$


13. Find the equation of the right bisector of the line segment joining the points $(3,4)$ and $(-1,2)$.

Ans: The right bisector of a line segment bisects the line segment at $90{}^\circ $.

 The end-point $A(3,4)$ and $B(-1,2)$of the line segment .

Accordingly, the mid-point of $AB=\left( \dfrac{3-1}{2},\dfrac{4+2}{0} \right)-(1,3)$ 

The Slope of $\text{AB}=-\dfrac{2-4}{-1-3}=\dfrac{-2}{-4}=\dfrac{1}{2}$

$\therefore $ The slope of the line perpendicular to $AB=-\dfrac{1}{\left( \dfrac{1}{2} \right)}=-2$ 

The equation of the line passing through $(1,3)$ and having a slope of $-2$ is $(y-3)=-2(x-1)$

$\Rightarrow y-3=-2x+2$

$\Rightarrow 2x+y=5$

Thus, the required equation of the line is $2x+y=5$.


14. Find the coordinates of the foot of the perpendicular from the points $(-1,3)$ to the line $3x-4y-16=0$

Ans: Let $(a,b)$ be the coordinates of the foot of the perpendicular from the points $(-1,3)$ to the line $3x-4y-16=0$.


(Image will be uploaded soon)


The slope of the line joining $(-1,3)$ and $(a,b)$, 

$\Rightarrow \text{  }$${{m}_{1}}=\dfrac{b-3}{a+1}$ 

Slope of the line $3x-4y-16=0$ or $y=\dfrac{3}{4}x-4,{{m}_{2}}=\dfrac{3}{4}$ ,

 The above two lines are perpendicular, at, ${{\text{m}}_{1}}{{\text{m}}_{2}}=-1$ 

$\therefore \Rightarrow \text{  }\left( \dfrac{b-3}{a+1} \right)\times \left( \dfrac{3}{4} \right)=-1$

$\Rightarrow \dfrac{3b-9}{4a+4}=-1$

$\Rightarrow 3b-9=-4a-4$

$\Rightarrow 4a+3b=5\quad .....(1)$

The point $(a$, b) lies on the line $3x-4y=16$.

$\therefore \Rightarrow \text{  }3a-4b=16.....(2)$

Solve the equations $(1)$ and (2), we get,

 $a=\dfrac{68}{25}$ and $b=-\dfrac{49}{25}$

Thus, the required coordinates of the foot of the perpendicular are $\left( \dfrac{68}{25},-\dfrac{49}{25} \right)$


15. The perpendicular from the origin to the fine $y=mx+c$ meets it at the point $(-1,2)$. Find the values of $m$ and $c$.

Ans: The  equation of the line is $y=mx+c$. 

The perpendicular from the origin meets the given line at $(-1,2)$. 

So, the line joining the points $(0,0)$ and $(-1,2)$ is perpendicular to the given line. 

$\therefore $ The slope of the line joining $(0,0)$ and $(-1,2)=\dfrac{2}{-1}=-2$ The slope of the given line is $\text{m}$.

 $\therefore \text{m}\times -2=-1\quad $     [The two  lines are perpendicular] 

$\Rightarrow m=\dfrac{1}{2}$.

The  points $(-1,2)$ lie on the given line, it satisfies the equation $y=mx+c$

$\therefore 2=\text{m}(-1)+c$

$\Rightarrow 2=2+\dfrac{1}{2}(-1)+c$

 $\Rightarrow c=2+\dfrac{1}{2}=\dfrac{5}{2}$

The respective values of $\text{m}$ and $\text{c}$ are $\dfrac{1}{2}$ and $\dfrac{5}{2}$.


16. If $p$ and $q$ are the lengths of the perpendicular from the origin to the lines $x\cos \theta -y\sin \theta =k$ $\cos 2\theta $ and $\text{x}\sec \theta +y$ cosec $\theta =\text{k}$, respectively, prove that ${{p}^{2}}+4{{q}^{2}}-{{k}^{2}}$

Ans: The equation of lines are $x\cos \theta -y\sin \theta =k\cos 2\theta \quad \ldots (1) $

$x\sec \theta +y$ cases $\theta =k\quad \ldots (2)$

The perpendicular distance (d) of a line $Ax+By+C=0$ from a point $\left( {{x}_{1}},{{x}_{2}} \right)$ is given by $d=\dfrac{\left| A{{x}_{1}}+B{{y}_{1}}+c \right|}{\sqrt{{{A}^{2}}+{{B}^{2}}}}$

Compare the  equation $\left( 1 \right)$  the general equation of line 

ie., $Ax+By+C=0$, we obtain $A=\cos \theta $ ,$B=\sin \theta $, and $C=k\cos 2\theta $.

$\text{p}$ is the length of the perpendicular from $(0,0)$ to line $\left( 1 \right)$. \[\therefore p=\dfrac{|A(0)+B(0)+C|}{\sqrt{{{A}^{2}}+{{B}^{2}}}}=\dfrac{|C|}{\sqrt{{{A}^{2}}+{{B}^{2}}}}\]

$\therefore p=\dfrac{|-k\cos 2\theta |}{\sqrt{{{\cos }^{2}}\theta +{{\sin }^{2}}\theta }}=|-k\cos 2\theta |\quad \ldots (3)$

 Compare the  equation (2) to the general equation of line ie., $Ax+By+C=0$, we obtain $A=\sec \theta $ , $B=\operatorname {cosec} \theta $, and $C=-k$

It is given that $q$ is the length of the perpendicular from $(0,0)$ to line (2). 

$\therefore \mathrm{q}=\frac{|\mathrm{A}(\mathrm{O})+\mathrm{B}(\mathrm{O})+\mathrm{C}|}{\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}}}=\frac{|\mathrm{C}|}{\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}}}=\frac{|-\mathrm{k}|}{\sqrt{\sec ^{2} \theta+\operatorname{cosec}^{2} \theta}}$

From (3) and (4), we have

$p^{2}+4 q^{2}-(|-k \cos 2 \theta|)^{2}+4\left(\frac{|-k|}{\sqrt{\sec ^{2} \theta+\operatorname{cosec}^{2} \theta}}\right)^{2}$

$=k^{2} \cos ^{2} 2 \theta+\frac{4 k^{2}}{\left(\sec ^{2} \theta+\operatorname{cosec}^{2} \theta\right)}$

$=k^{2} \cos ^{2} 2 \theta+\frac{4 k^{2}}{\left(\frac{1}{\cos ^{2} \theta}+\frac{1}{\sin ^{2} \theta}\right)}$

$=k^{2} \cos ^{2} 2 \theta+\frac{4 k^{2}}{\left(\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin ^{2} \theta \cos ^{2} \theta}\right)}$

$=k^{2} \cos ^{2} 2 \theta+\frac{4 k^{2}}{\left(\frac{1}{\sin ^{2} \theta \cos ^{2} \theta}\right)}$

$=k^{2} \cos ^{2} 2 \theta+4 k^{2} \sin ^{2} \theta \cos ^{2} \theta$

$=k^{2} \cos ^{2} 2 \theta+k^{2}(2 \sin \theta \cos \theta)^{2}$

$=k^{2} \cos ^{2} 2 \theta+k^{2} \sin ^{2} 2 \theta$

$=k^{2}\left(\cos ^{2} 2 \theta+\sin ^{2} 2 \theta\right)$

$=\mathbf{k}^{2}$

Hence, vre proved that $\mathrm{p}^{2}+4 \mathrm{q}^{2}=\mathrm{k}^{2}$


17. In the triangle $\text{ABC}$ with vertices $\text{A}(2,3),\text{B}(4,-1)$ and $\text{C}(1,2)$, find the equation and length of altitude from the vertex $A$.

Ans: Let $\text{AD}$ be the altitude of the triangle $\text{ABC}$ from the vertex $\text{A}$. 

 $\Rightarrow \text{  }AD\bot BC$,


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The slope of the line BC $\dfrac[21][1-4]$=$-1$

Therefore slope of line AD  $= \dfrac{-1}{-1}$=$1$

The equation of the line AD passing through the point $A (2,3)$ and having a slope $1$ is $(y-3)=1(x-2)$

$\Rightarrow x-y+1=0$

$\Rightarrow y-x=1$

Therefore, equation of the altitude from a vertex$A=y-x=1$.

Length of $AD=$ Length of the perpendicular from A $(2,3)$ to$BC$.

The equation of $\text{BC}$ is 

$(y+1)=\dfrac{2+1}{1-4}(x-4)$

$\Rightarrow \text{  }(y+1)=-1(x-4)$

$\Rightarrow y+1=-x+4$

$\Rightarrow x+y-3=0$$\ldots (1)$

The perpendicular distance $(d)$ of a line $Ax+By+C=0$ from a point $\left( {{x}_{1}},{{y}_{1}} \right)$ is $d=\dfrac{\left| A{{x}_{1}}+B{{y}_{1}}+c \right|}{\sqrt{{{A}^{2}}+{{B}^{2}}}}$

 Compare the equation $(1)$ to the general equation of the line$Ax+By+C=0$.

We get,

 $A=1$,$\text{B}=1$ and $C=-3$

Length of AD $=  \dfrac{\left| 1\times 2+1\times 3-3 \right|}{\sqrt{{{1}^{2}}+{{1}^{2}}}}$ units 

$=\dfrac{\left| 2 \right|}{\sqrt{2}}$units $=\sqrt{2 }$units.

Length of AD $=\sqrt{2 }$units.

Thus, the equation and length of the altitude from vertex $A$ are $y-x=1$ and $\sqrt{2}$ wits respectively.


18. If $\text{p}$ is the length of the perpendicular from the origin to the line whose intercepts on the axes are $a$, and $\text{b}$, then show that $\dfrac{1}{{{p}^{2}}}=\dfrac{1}{{{a}^{2}}}+\dfrac{1}{{{b}^{2}}}$.

Ans: The equation of a line whose intercepts on the axes are $a$ and $\text{b}$ is $\dfrac{x}{a}+\dfrac{y}{b}=1$

$\text{bx}+ay=\text{ab}$

Or $bx+ay-ab=0$

The perpendicular distance $(d)$ of a line $Ax+By+C=0$ from a point $\left( {{x}_{1}},{{y}_{1}} \right)$ is $d=\dfrac{\left| A{{x}_{1}}+B{{y}_{1}}+c \right|}{\sqrt{{{A}^{2}}+{{B}^{2}}}}$

Compare the equation $(1)$ with the general equation of the line$Ax+By+C=0$, 

We obtain$A=b$,$\text{B}=a$ and $C=-ab$

Therefore, if $p$ is the length of the perpendicular from a point $\left( {{x}_{1}},{{y}_{1}} \right)=(0,0)$ to lime$\left( 1 \right)$. We obtain $p  =  \dfrac{\left| A\left( 0 \right)  +  B\left( 0 \right)  -  ab \right|}{\sqrt{{{b}^{2}}+{{a}^{2}}}}$

$\Rightarrow p=\dfrac{|-ab|}{\sqrt{{{b}^{2}}+{{a}^{2}}}}$.

Square both sides, 

We get

$\Rightarrow \text{  }{{p}^{2}}  =  \dfrac{{{\left( -  ab \right)}^{2}}}{{{a}^{2}}  +  {{b}^{2}}}$

$\Rightarrow \text{  }{{p}^{2}}\left( {{a}^{2}}  +  {{b}^{2}} \right)  =  {{a}^{2}}{{b}^{2}}$

$\Rightarrow \text{  }\dfrac{{{a}^{2}}  +  {{b}^{2}}}{{{a}^{2}}{{b}^{2}}}  =  \dfrac{1}{{{p}^{2}}}$

$\Rightarrow \text{  }\dfrac{1}{{{p}^{2}}}  =  \dfrac{1}{{{a}^{2}}}  +  \dfrac{1}{{{b}^{2}}}$

Hence proved that  $\dfrac{1}{{{p}^{2}}}=\dfrac{1}{{{a}^{2}}}+\dfrac{1}{{{b}^{2}}}$


NCERT Solutions for Class 11 Maths Chapters

 

NCERT Solution Class 11 Maths of Chapter 10 All Exercises

Chapter 10 - Straight Lines Exercises in PDF Format

Exercise 10.1

14 Questions & Solutions

Exercise 10.2

20 Questions & Solutions

Exercise 10.3

18 Questions & Solutions

Miscellaneous Exercise

24 Questions & Solutions

Topics Covered in Excercise 10.3 NCERT

Exercise 10.3 of Straight Lines is based on the following topics:

1. General Equation of a straight line

The formula Ax + By + C = 0 is said to be the 'general form' for the equation of a straight line, where A, B, and C are any three real numbers. 


2. Different Forms of Straight Line

  • General form

  • Standard form

  • Slope-Intercept form

  • Point-Slope form

  • Intercept form


3. The Distance of a Point From a Line

Here, the distance from a point to a line is the shortest distance from a given point to a point on an infinite straight line. This shortest distance is the perpendicular distance of the point to the line.


4. Distance Between Two Parallel Lines

The distance between two parallel lines is the perpendicular distance which is the minimum distance from any point on one line to the other line.


NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Exercise 10.3

Opting for the NCERT solutions for Ex 10.3 Class 11 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 10.3 Class 11 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 11 students who are thorough with all the concepts from the Subject Straight Lines textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 11 Maths Chapter 10 Exercise 10.3 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 11 Maths Chapter 10 Exercise 10.3, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

Do not delay any more. Download the NCERT solutions for Class 11 Maths Chapter 10 Exercise 10.3 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.

FAQs on NCERT Solutions for Class 11 Maths Chapter 10: Straight Lines - Exercise 10.3

1. What are the topics and subtopics of Class 11 Maths Chapter 10- Straight Lines?

The topics and subtopics of Class 11 Maths Chapter 10 titled Straight Lines are given below. Take a look.

  • 10 - Straight Lines

  • 10.1 - Introduction

  • 10.2 - Slope of Line

  • 10.3 - Various Forms of the Equation of Line

  • 10.4 - General Equation of Line

  • 10.5 - Distance of a Point From a Line

2. What does Exercise 10.3 of NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines deal with?

NCERT Solutions are provided to help the students in understanding the steps to solve mathematical problems that are provided in the textbook. The subject matter experts at Vedantu stick to the syllabus while preparing the solutions. The problem-solving method provided in the examples is followed while preparing the NCERT Solutions for class 11 as well.


The Exercise 10.3 of NCERT Solutions for Class 11 Maths Chapter 10- Straight Lines is based on the following topics:

  1. General Equation of a Line

    1. Different forms of Ax + By + C = 0

  2. Distance of a Point From a Line

    1. Distance between two parallel lines

3. How many questions are there in the Exercise 10.3 of NCERT Solutions for Class 11 Maths Chapter 10 - Straight Lines?

18 questions are there in total in the Exercise 10.3 of NCERT Solutions for Class 11 Maths Chapter 10 - Straight Lines.

4. Can I get access to the PDF format of the Exercise 10.3 of NCERT Solutions for Class 11 Maths Chapter 10 - Straight Lines?

Free PDF download of NCERT Solutions for Class 11 Maths Chapter 10 Exercise 10.3 (Ex 10.3) is available on Vedantu site along with all chapter exercises at one place. These solutions are prepared by the expert teachers as per CBSE curriculum and guidelines. Class 11 Maths Chapter 10 Straight Lines Exercise 10.3 questions with solutions to help you to revise the entire syllabus and score the best possible marks.

5. What all things are included in Class 11 Maths Exercise 10.3?

Chapter 10 Exercise 10.3 NCERT Solutions for Class 11 Maths Straight Lines explains how algebraic equations can be used to learn about a straight line's trajectory. There are numerous possibilities for a straight line, such as being parallel to the X or Y axis or going through the origin. When we look at how the equation of such a line is simplified when some of the terms in the equation become zero, we can identify some of these circumstances that make the straight line a little more peculiar.

6. How does Algebra provide insight into the concepts of a straight line?

Simply looking at the simplified equation y= MX or y=0 or x=0 gives us an instant knowledge of the line's specialization. The questions in this NCERT Solutions Class 11 Mathematics Chapter 10 exercise 10.3 show how Algebra may help us understand the principles of a straight line by using its various forms. Taking notes from Vedantu will help students in solving these questions easily. Then some questions use line equations to assist us to discover the relationship between a group of lines.

7. How can you find perpendicular and parallel lines?

We can determine if they are parallel or perpendicular and if they intersect, where they intersect. Some questions incorporate trigonometry, and we can see how by combining different concepts, we may create an outstanding tool for analysis. In Class 11 Maths NCERT Solutions In Chapter 10, Exercise 10.3, there are 18 questions, seven of which are brief and eleven of which are extended answer-type questions. Vedantu provides every concept in the form of a PDF too, which you can download free of cost.

8. How can students solve the 10.3 Exercise?

Class 11 Maths NCERT Solutions Chapter 10 Exercise 10.3 Straight Lines focuses on utilizing various slope shapes and identifying key elements associated with them. Thus, memorizing the formula for the general form of slopes and understanding how to apply it to estimate various characteristics of a straight line is the key to successfully answering these issues. Vedantu app and the website has all the questions on its website in which you will find all the solutions of the exercise present in the NCERT book.

9. How can imagining different values help in solving this exercise easily?

The inclination should be to read the problem and visualize how it will be plotted on the axis when doing the NCERT solutions Class 11 Maths Chapter 10 Exercise 10.3 Straight Lines. Visualizing how it will behave for various values of the variables while focusing on the unique conditions might also aid in the learning process. On Vedantu, you can find the basic concepts of the chapter which will help you in understanding the topic in depth. Solving numerical will be easier if you know the concept behind it.