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We define a straight line as a curve where every point on the line segment joining any two points lies on it.

Let’s throw light on the coordinate geometry to prove that every first-degree equation in x, y represents a straight line.

Let ax + by + c = 0 be a first-degree equation in x,y where a, b, and c are constant.

Let points P (x1, y1) and Q (x2 , y2 ) be any two points on the curve represented by ax + by + c = 0.

Then, ax1+ by1 + c = 0, and ax2 + by2 + c = 0 ….(1)

Let M be the point on the line segment joining P and Q. Let’s say this point M divides PQ in the ratio of n:1. Then the coordinates of M are:

\[\frac{(nx2 + x1)}{n + 1}\] , \[\frac{(ny2+ y1)}{n + 1}\]

Now putting in the equation ax + by + c = 0, we have,

= a \[\frac{(nx2 + x1)}{n + 1}\] + b \[\frac{(ny2+ y1)}{n + 1}\] + c

= \[\frac{n [(ax2 +by2 + c) + (ax1+ by1 + c )]}{n + 1}\]

From equation (1), we get,

= n x 0 + 0

= 0

This means point M [ \[\frac{(nx2 + x1)}{n + 1}\] , \[\frac{(ny2+ y1)}{n + 1}\] ] lies on the curve represented by ax + by + c = 0.

Thus, every point on the line segment joins P and Q lines on ax + by + c = 0.

Hence, ax + by + c = 0 corresponds to a straight line.

The general equation of a straight line y = m * x + c.

m is the gradient (slope) of the line

We define the slope of a line as the trigonometrical tangent of the angle that a line makes with the positive direction of the x-axis in an anticlockwise sense.

c = intercept on the y-axis, and

∴ y = c is the value where the line cuts the y-axis.

Now, let’s look at an example to understand this equation of a straight line.

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∠MPQ = Ө

∴ The slope of a line l = m = tan Ө = QM/PM = \[\frac{(y2 - y1)}{(x2 - x1)}\]

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For example,

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m = \[\frac{(y2 - y1)}{(x2 - x1)}\] = \[\frac{(25 - 5)}{(30 - 15)}\] = 4/3

Hence, y = m * x + c is an equation of straight, which is also known as a slope-intercept form.

Let’s take two different straight lines intersecting at the origin, and the equation of each is:

L1 = ax1+ by1 +c, and L2 = ax2 + by2 + c = 0

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Now, let’s take any point P on the line L1.

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The path it travels on these two lines is its locus. Now, we need to find the focus of this point that can travel on both these lines. This means coordinates of P (h,k) should satisfy an equation that shows it can be on L1 and L2.

Therefore, to find the equation of a pair of straight lines, we multiply the equations of L1 and L2, i.e.,

(ax1+ by1 +c) * (ax2 + by2 + c) = 0.

Since these two lines intersect at the origin, this equation is homogeneous.

If point P lies on line L1, then the product is zero, and if it is on L2, then also it is zero.

So, the point that satisfies the equation or not can be determined by the overall product, which should always be zero.

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We have two equations y = m1x, and y = m2x

Equation of pair of straight lines = (m1x - y) (m2x - y) = m1m2x^{2} - (m1 + m2)xy + y^{2} , which is a homogeneous 2nd degree equation.

If the equations y = m1x, and y = m2x are represented by , \[ax^2 + 2hxy + by^2 = 0\] then

m1 + m2 = - 2h/b, and m1m2 = a/b

From the equation , \[(a)x^2 + (2hy)x + by^2 = 0\] we can consider an equation to be quadratic to x, then we have:

x = \[\frac{-2hy ± \sqrt{4h^2y^2- 4 aby^2}}{2a}\]

Here, we will get two lines on because we have a +, and a - sign.

1. We know that angle between two lines is |tan Ө = ± \[\frac{m2 -m1}{1 + m1m2}\] |, and on substituting the value of m1 + m2 = - 2h/b, and m1m2 = a/b, we get, 2

Ө = tan^{-1 }[|\[\frac{2\sqrt{2 h^2- ab}}{(a + b)}\]|], which is the angle between the two lines expressed as \[ax^2 + 2hxy + by^2 = 0\]

2. If these two lines are parallel, then the angle between them, i.e., Ө = 0°, then

|tan Ө = ± \[\frac{m2 -m1}{1 + m1m2}\] | = 0 ⇒ m1 = m2

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This means when two lines are parallel; their slopes are equal.

3. If two lines are perpendicular to each other, the angle between them, i.e., Ө = 90°, then

1/tan 90° = 1/∞ = |\[\frac{1 + m1m2}{m2 - m1}\]| = 0 ⇒ m1 = m2 = - 1

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This means when two lines are perpendicular to each other, the product of their slopes is -1, i.e., if m is the slope of L1, then the slope L2 ⏊ to it is (-1/m).

FAQ (Frequently Asked Questions)

Q1: Find the equation of a straight line which makes an angle of tan^{-}√3 with the x-axis, and cuts off an intercept of √3/2 with the y-axis.

Ans: Here, m = tan Ө = tan (tan^{-}√3) = √3

The equation in y = mx +c form is:

y = √3x + √3/2

Q2: The ⏊_{PERPENDICULAR} from the origin (0,0) to a line meets at the point (- 3, 7). Find the equation of the line.

Ans: We have a slope, m = (y2 - y1)/(x2 - x1) = (7 - 0)/(- 3 - 0)= -7/3

Slope of a line perpendicular to this line = - 1/m = 3/7

(y - 7) = 3/7(x + 3)

Required equation=3x - 7y +52 = 0

Q3: If the pairs of lines 2x^{2} - 3xy + qy^{2}=0,and 4x^{2} - xy -3y^{2}=0 have one line in common, then q =?

Ans: We have L_{1} = 4x^{2} - xy -3y^{2}=0 , and L_{2 }= 2x^{2} - 3xy + qy^{2}=0

Now, let’s look at L_{1},

4x^{2} - 4xy + 3xy -3y^{2}=0

4x (x - y) + 3y (x - y) = 0

= (x-y)(4x + 3y) = 0

So, x - y = 0 ⇒ y/x = 1..(1), and

4x + 3y = 0 ⇒ y/x = - 4/3…(2)

For L_{2},

2x^{2 }- 3xy + qy^{2}=0 dividing by x^{2} , we get,

2 - 3(y/x) + q(y/x)^{2}= 0…(3)

Case1: Substituting the value of y/x from eq(1) in (3)

2 - 3(1) + q(1)^{2} = 0, we get,

q = 1

Case2: Substituting the value of y/x from eq(2) in (3)

2 - 3(4/3) + q(4/3)^{2} = 0, we get,

q = 9/8

So, the value of q are 1, 9/8 |

Q4: The pair of lines represented by 2c^{2}+ 5xy + (c^{2} - 3)y^{2} = 0 are perpendicular to each other for how many values of c?

Ans: We know the 2nd-degree equation, i.e., Ax^{2} + 2Hxy + By^{2}=0

For a perpendicular condition, make A + B = 0,

2c + (c^{2} - 3) = 0 = 0 ⇒ c^{2}+ 2c -3= 0

Using quadratic formula,

-2 ± √4 + 12/ 2

On solving, we get two values of c, i.e., 1 and - 3.