# HOW TO FIND THE INTERCEPTS MADE BY THE EQUATION OF THE LINE

Question 1 :

Find the intercepts made by the following lines on the coordinate axes.

(i) 3x − 2y − 6 = 0

Solution :

3x − 2y − 6 = 0

3x − 2y  = 6

Divide the entire equation by 6, we get

(3x/6) - (2y/6)  =  6/6

(x/2) + y/(-3)  =  1

x-intercept  =  2

y - intercept  =  -3

(ii) 4x + 3y +12 = 0

4x + 3y + 12 = 0

4x + 3y  = -12

Divide the entire equation by -12, we get

(4x/(-12)) + (3y/(-12))  =  -12/(-12)

x/(-3) + y/(-4)  =  1

x-intercept  =  -3

y - intercept  =  -4

Question 2 :

Find the equation of a straight line

(i) passing through (1,-4) and has intercepts which are in the ratio 2 : 5

Solution :

x-intercept  =  2k

y-intercept  =  5k

Equation of the line when intercepts are given

(x/a) + (y/b)  =  1

The required line is passing through the point (1, -4)

(1/2k) + (-4/5k)  =  1

(5 - 8)/10k  =  1

-3/10k  =  1

10k  =  -3

k  =  -3/10

a  =  2(-3/10)  =  -3/5

b = 5(-3/10)  =  -3/2

Equation of the line :

x/(-3/5) + y/(-3/2)  =  1

-5x - 2y  = 3

5x + 2y + 3  =  0

(ii) passing through (-8, 4) and making equal intercepts on the coordinate axes

Solution :

a = b (makes equal intercepts)

Equation of the line :

x/a + y/b  =  1

-8/a + 4/a  =  1

(-8 + 4)/a  =  1

a  =  -4

Equation of the required line :

x/(-4) + y/(-4)  =  1

x + y  =  -4

x + y + 4  =  0

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