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

Answer:

$y = -\frac{1}{3}x + \frac{8}{3}$

Step-by-step 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:

$\frac{x}{a} + \frac{y}{b} = 1$

Where

b = Intercept on y axis.

a = Intercept on x axis.

From the question:

$a = 3b$

The equation becomes:

$\frac{x}{3b} + \frac{y}{b} = 1$

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

$b = \frac{8}{3}$

Substitute $b = \frac{8}{3}$ in $x + 3y = 3b$ to get the equation

$x + 3y = 3 * \frac{8}{3}$

$x + 3y = 8$

Make 3y the subject

$3y = -x + 8$

Make y the subject

$y = -\frac{1}{3}x + \frac{8}{3}$