Question

$1)$ Equation of a line is
$3x-4y+10=0$. Find its
$i)$ slope

$2)$ Equation of a line is
$3x-4y+10=0$. Find its
$ii)$ $x$-and $y$-intercepts

Answer 1

$1)$ Given $3x-4y+10=0\Rightarrow 4y=3x+10$

$\Rightarrow y=\frac{3}{4}x+\frac{10}{4}$

$\Rightarrow y=\frac{3}{4}x+\frac{5}{2}$

The above equation is of the form

$y=mx+c$

Hence,

slope $m=\frac{3}{4},$

$2)$ Finding $x\&y$-intercepts

Given: $3x-4y+10=0$

$\Rightarrow 3x-4y=-10$

Dividing both sides by $-10$

$\Rightarrow \frac{3x-4y}{-10}=\frac{-10}{-10}$

$\Rightarrow \frac{3x}{-10}-\frac{4y}{-10}=1$

$\Rightarrow \frac{3x}{-10}+\frac{4y}{10}=1$

$\Rightarrow \frac{x}{\frac{-10}{3}}+\frac{y}{\frac{5}{2}}=1$

The above equation of the form $\frac{x}{a}+\frac{y}{b}=1$

hence, $a=x-\text{intercept}=\frac{-10}{3}$

and $b=y-\text{intercept}=\frac{5}{2}$