Question

Transform the equation $4x-3y+12=0$ into
slope intercept form
Intercept form
Normal form

Answer 1

$4x-3y+13=0$

$\left(1\right)$ slope intercept from

$4x+12=3y$

$y=\frac{4}{3}x+\frac{12}{3}\Rightarrow \frac{4}{3}x+4$

$\left(2\right)$ Intercept from

[from $\left(1\right)$] which intercept at $y-$axis

equation for $x$ which intercept at $x-$axis

$4x=3y-12$

$n=\frac{3}{4}y-\frac{-12}{4}\Rightarrow \frac{3}{4}y-3$

Intercept at $x-$axis $\Rightarrow x=\frac{3}{4}y-3$

Intercept at $y-$axis $\Rightarrow y=\frac{4}{3}x+4$

$\left(3\right)$ Normal form

$4x-3y+12=0$

$\Rightarrow \frac{4x}{3}-\frac{3y}{5}+\frac{12}{5}=0$

$\Rightarrow \frac{4x}{5}-\frac{3y}{5}=\frac{-12}{5}$

Dividing by

$[\sqrt{4^2+3^2}\Rightarrow \sqrt{16+9}\Rightarrow \sqrt{25}=5]$

is the normal form