Draw the graph of the pair of equations $ 2 x+y=4 $ and $ 2 x-y=4 $. Write the vertices of the triangle formed by these lines and the $ y $-axis. Also find the area of this triangle.

Given:

The equations of two of the sides of the given triangle are:

\( 2 x+y=4 \) and \( 2 x-y=4 \)

To do:

We have to determine the vertices and the area of the given triangle formed by these lines and the \( y \)-axis.

Solution:

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

For equation $2x+y=4$,

$y=4-2x$

If $x=0$ then $y=4-2(0)=4$

If $x=2$ then $y=4-2(2)=4-4=0$

For equation $2x-y=4$,

$y=2x-4$

If $x=0$ then $y=2(0)-4=-4$

If $x=2$ then $y=2(2)-4=4-4=0$

Equation of y-axis is $x=0$. The above situation can be plotted graphically as below:

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

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

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

In the graph, the height of the triangle is the distance between point $(2,0)$ and y-axis and the length of the base is the distance between $(0, 4)$ and $(0, -4)$

Height of the triangle$=2$ units.

Length of the base of the triangle$=4+4=8$ units.

Area of the triangle formed by the given lines $=\frac{1}{2}\times2\times8$

$=8$ sq. units. 

The area of the triangle is $8$ sq. units.

Tutorialspoint

Updated on: 10-Oct-2022

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