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

y-intercept (b) =   2t

(3/3t)+(4/2t)  =  1

(1/t)+(2/t)  =  1

3/t  =  1

3  =  t

t  =  3

So, x-intercept (a)  =  3t  =  3(3) ==> 9

y-intercept (b)  =  2t  =  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

t  =  8

By applying the value of t in (1), we get

(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 :

(x/a) + (y/b)  =  1

Sum of intercept  =  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

a²-9a+18  =  0

(a-3) (a-6)  =  0

a  =  3 and a  =  6

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 required equations are

2x+y-6  =  0 or x+2y-6  =  0

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