# EQUATION OF LINE WITH A POINT AND RATIO OF INTERCEPT IS GIVEN

Example 1 :

Find the equation of the straight line passing through the point (3, 4) and has intercepts which are in the ratio 3:2.

Solution :

Intercept form :

x/a + y/b = 1

Intercepts are in the ratio 3:2

So, x-intercept (a) = 3t and y-intercept (b) = 2t.

3/3t + 4/2t = 1

1/t + 2/t = 1

3/t = 1

3 = t

So,

x-intercept (a) = 3(3) = 9

y-intercept (b) = 2(3) = 6

x/9 + y/6 = 1

(2x +3y)/18 = 1

2x + 3y = 18

2x + 3y - 18 = 0

Example 2 :

Find the equation of the straight line passing through the point (5, -3) and whose intercepts on the axes are equal in magnitude but opposite in sign.

Solution :

Intercept form :

x/a + y/b = 1

Intercepts are equal in magnitude but opposite in sign.

x-intercept (a) = t

y-intercept (b) = -t

The required line is passing through the point (5, -3)

5/t + -3/(-t) = 1 ----(1)

5/t + 3/t = 1

(5 + 3)/t = 1

8/t = 1

8 = t

Substitute t = 8 in (1).

x/8 + y/(-8) = 1

(x - y)/8 = 1

x - y = 8

x - y - 8 = 0

Example 3  :

Find the equation of the straight lines passing through the point (2, 2) and sum of the intercepts is 9.

Solution :

Intercept form equation of a line :

x/a + y/b = 1

Here, a and b are x and y-intercepts respectively.

Sum of the intercepts = 9.

a + b = 9

b = 9 - a

The required line is passing through the point (2, 2).

2/a + 2/(9 - a) = 1

[2(9 - a) + 2a]/[a(9 - a)] = 1

[18 - 2a + 2a]/(9a - a2) = 1

18/(9a - a2) = 1

18 = 9a - a2

a- 9a + 18 = 0

(a - 3)(a - 6) = 0

a = 3 and a = 6

 If a = 3,b = 9 - 3b = 6 If a = 6,b = 9 - 6b = 3

Equation of the line when a = 3 and b = 6 :

x/3 + y/6 = 1

(2x + y)/6 = 1

2x + y = 6

2x + y - 6 = 0

Equation of the line when a = 6 and b = 3 :

x/6 + y/3 = 1

(x + 2y)/6 = 1

x + 2y = 6

x + 2y - 6 = 0

So, the equation of the required line is

2x + y - 6 = 0 or x + 2y - 6 = 0

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