Question

Find the equation of the straight line which passes through the point (3,-2) and cuts off positive intercepts on the x and y axis which are in the ratio 4:3.​

Answer 1

Answer:

3 x + 4 y - 1 = 0

Step-by-step explanation:

Let the unit constant 'k' for constant.

a' = 4 a and c = 3 a

We have slope intercept form :

x / a' + y / b = 1

Passing point is given ( 3 , - 2 ) :

3 / 4 a - 2 / 3 a = 1

9 - 8 = 12 a

a = 1 / 12

Now we have value of both intercept :

a' = 4 / 12 = 1 / 3

c = 3 / 12 = 1 / 4

Equation of line :

x / a' + y / b = 1

x / 1 / 3 + y / 1 / 4 = 1

3 x + 4 y - 1 = 0

Therefore , equation of line is 3 x + 4 y - 1 = 0.