Question

Find the equation of the straight line which passes through the point (-1,3) and make intercept on x - axis as thrice that on y - axis .​

Answer 1

`y = -(1)/(3)x + (8)/(3)`

Explanation:

Given

`(x_1,y_1) = (-1,3)`

`Intercept\ on\ x-axis = Intercept\ on\ y-axis.`

Required

Determine the equation

First, we calculate the intercepts using:

`(x)/(a) + (y)/(b) = 1`

Where

b = Intercept on y axis.

a = Intercept on x axis.

From the question:

`a = 3b`

The equation becomes:

`(x)/(3b) + (y)/(b) = 1`

Multiply through by 3b

`x + 3y = 3b`

We have:
`(x_1,y_1) = (-1,3)`

So:

`-1 + 3 * 3 = 3b`

`8 = 3b`

Make b the subject

`b = (8)/(3)`

Substitute
`b = (8)/(3)` in
`x + 3y = 3b` to get the equation

`x + 3y = 3 * (8)/(3)`

`x + 3y = 8`

Make 3y the subject

`3y = -x + 8`

Make y the subject

`y = -(1)/(3)x + (8)/(3)`