Question

Graphically, solve the following pair of equations: $2x+y=6,2x-y+2=0$ Find the ratio of the areas of the two triangles formed by the lines representing these equations with the $x-$axis and the lines with the $y-$axis.

Answer 1

Step 1: Evaluate the first equation

The given equation is $2x+y=6$.

Assuming various values of $x$ and $y$ we get the following table:

$x$	$0$	$1$	$3$
$y$	$6$	$4$	$0$

Step 2: Evaluate the second equation

The given equation is $2x-y+2=0$.

Assuming various values of $x$ and $y$ we get the following table:

$x$	$0$	$1$	$-1$
$y$	$2$	$4$	$0$

Step 3: Form the Diagram

Step 4: Evaluate the required ratio

Let $A_1$ and $A_2$ represent the areas of triangles $ACE$ and $BDE$ respectively.

Let, Area of triangle formed with $x-$axis$=A_1$

$A_1=$ Area of $∆ACE$

$=\frac{1}{2}\times AC\times PE$

$=\frac{1}{2}\times 4\times 4$

And Area of triangle formed with $y-$ axis $=A_2$

$A_2=$ Area of $∆BDE$

$=\frac{1}{2}\times BD\times QE$

$=\frac{1}{2}\times 4\times 1$

$\therefore A_1:A_2=\frac{\frac{1}{2}\times 4\times 4}{\frac{1}{2}\times 4\times 1}$

$=4:1$

Hence, the ratio of the areas of the triangle is $4:1$.