Question

Draw the graph of the straight line given by the equation $4x-3y+36=0$. Calculate the area of the triangle formed by the line drawn and co-ordinate axes.

Answer 1

$4x-3y+36=0$
$\Rightarrow 4x-3y=-36$
$\Rightarrow -3y=-36-4x$
$\Rightarrow 3y=36+4x$
$\Rightarrow y=\frac{36+4x}{3}$
When $x=-6,y=\frac{36+4\times (-6)}{3}=\frac{36-24}{3}=4$
When $x=-3,y=\frac{36+4\times (-3)}{3}=\frac{36-12}{3}=8$

When $x=-9,y=\frac{36+4\times (-9)}{3}=\frac{36-36}{3}=0$

$x$	$-9$	$-3$	$-6$
$y$	$0$	$8$	$4$

Hence, the straight line cuts the co-ordinate axis at $A(0,12)$ and $B(-9,0)$.
$\therefore$ The triangle $△AOB$ is formed.
Area of the triangle $AOB=\frac{1}{2}\times AO\times OB$
$=\frac{1}{2}\times 12\times 9$
$=54sq.units$
$\therefore$ Area of the triangle is $54sq.units$.