Question

Equation of a line is $3x-4y+10=0$. Find its
$\left(i\right)$ slope,
$\left(ii\right)x-$ and $y$- intercepts.

Answer 1

given equation of line is

$3x-4y+10=0$

rewriting in general form

$4y=3x+10$

or

$y=\frac{3}{4}x+\frac{10}{4}\dots \left(1\right)$

and we know that general form a straight line is

$y=mx+c\dots \left(2\right)$

where $m$ is slope of the line and $c$ is the y-intercept of the line

hence, by comparing eq(1) and eq(2) we get

slope of line $m=\frac{3}{4}$ (slope of line)

and y-intercept is defined as value of y at x=0

hence, from equation (1)

$y=\frac{3}{4}\times 0+\frac{10}{4}$

or $y=\frac{10}{4}$ (y-intercept)

and x-intercept is defined as value of x at y=0

hence, from equation (1) we get

$0=\frac{3}{4}x+\frac{10}{4}$

$3x=10$

or $x=-\frac{10}{3}$ (x-intercept)