Draw the graphs of the following equations:

$2x\ -\ 3y\ +\ 6\ =\ 0$
$2x\ +\ 3y\ -\ 18\ =\ 0$
$y\ -\ 2\ =\ 0$
Find the vertices of the triangle so obtained. Also, find the area of the triangle.

Given:

The equations of the sides of the given triangle are:

$2x\ -\ 3y\ +\ 6\ =\ 0$

$2x\ +\ 3y\ -\ 18\ =\ 0$

$y\ -\ 2\ =\ 0$

To do:

We have to determine the vertices and the area of the given triangle.

Solution:

To represent the above equations graphically we need at least two solutions for each of the equations.

For equation $2x-3y+6=0$,

$2x=3y-6$

$x=\frac{3y-6}{2}$

If $y=0$ then $x=\frac{3(0)-6}{2}=\frac{-6}{2}=-3$

If $y=2$ then $x=\frac{3(2)-6}{2}=\frac{6-6}{2}=0$

$x$	$-3$	$0$
$y$	$0$	$2$

For equation $2x+3y-18=0$,

$2x=18-3y$

$x=\frac{18-3y}{2}$

If $y=6$ then $x=\frac{18-3(6)}{2}=\frac{18-18}{2}=0$

If $y=4$ then $x=\frac{18-3(4)}{2}=\frac{18-12}{2}=\frac{6}{2}=3$

$x$	$0$	$3$
$y$	$6$	$4$

For equation $y-2=0$,

$y=2$

This implies, for every value of x, $y=2$

$x$	$0$	$6$
$y$	$2$	$2$

The above situation can be plotted graphically as below:

The lines AB, CD and EF represent the equations $2x-3y+6=0$, $2x+3y-18=0$ and $y-2=0$ respectively.

As we can see, the points of intersection of the lines AB, CD and EF taken in pairs are the vertices of the given triangle.

Hence, the vertices of the given triangle are $(3,4), (6,2)$ and $(0,2)$.

 We know that,

Area of a triangle$=\frac{1}{2}bh$

In the graph, the height of the triangle is the distance between point D and EF.

Height of the triangle$=4-2=2$ units.

Base of the triangle$=$Distance between the points B and F.

Base of the triangle$=6$ units.

Area of the triangle formed by the given lines $=\frac{1}{2}\times2\times6$

$=6$ sq. units. 

The area of the triangle is $6$ sq. units.