Question

Transform the equation $3x+4y+12=0$ in to normal form.

Answer 1

We have,

$3x+4y+12=0$

$3x+4y=-12$

Now,

$=\sqrt{coefficientof(x{)}^2+coefficientof(y{)}^2}$

$=\sqrt{(3{)}^2+(4{)}^2}$

$=\sqrt{9+16}$

$=\sqrt{25}$

$=5$

Now Divide by $5$ on both side,

$=\left(\frac{3}{5}\right)x+\left(\frac{4}{5}\right)y=\frac{-12}{5}$