WRITE THE EQUATION OF A LINE IN STANDARD FORM WITH ONE POINT AND SLOPE

Equation of the line using slope point form :

(y - y1)  =  m(x - x1)

Here we have to consider the given point as (x1, y1) and slope as 'm'.

Example 1 :

Find the equation of the straight line whose.

(i)  Slope is -4 and passing through (1, 2).

(ii)  Slope is 2/3 and passing through (5, -4).

Solution :

Part (i) :

To find the equation of the straight line, we have to use the formula of slope point form.

Slope (m) = -4, x1 = 1 and y1 = 2.

Equation of the line  :

(y - y1) = m(x - x1)

(y – 2) = -4(x – 1)

y – 2 = -4x + 4

4x + y – 2 - 4 = 0

4x + y - 6 = 0

Part (ii) :

To find the equation of the straight line we have to use the formula of slope point form.

Slope (m) = 2/3, x1 = 5 and y1 = -4.

Equation of the line  :

(y - y1) = m(x - x1)

y – (-4) = (2/3)(x – 5)

 (y + 4) = (2/3)(x – 5)

3(y + 4) = 2(x – 5)

3y + 12 = 2x – 10

2x – 3y - 10 - 12 = 0

2x – 3y – 22 = 0

Example 2 :

Find the equation of the straight line which passes through the midpoint of the line segment joining

(4, 2) and (3, 1)

whose angle of inclination is 30 degree.

Solution :

First we have to find midpoint of the line segment joining the points (4, 2) and (3, 1).

Midpoint = (x1 + x2)/2, (y1 + y2)/2

= (4 + 3)/2, (2 + 1)/2

= (7/2, 3/2)

angle of inclination = 30°.

θ = 30°

Slope (m) = tan θ

m = tan 30°

m = 1/√3

Equation of the line :

y - y1 = m(x - x1)

y - 3/2 = (1/√3)(x - 7/2)

2y - 3 = (1/√3)(2x - 7)

2√3y - 3√3 = 2x - 7

2x - 2√3y - 7 + 3√3 = 0

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