Question

draw the graph of x+y=12 and x-y=12 on the same graph,also calculate the area of triangle formed by these two lines and y axis



please plot the graph​

Answer 1

Answer:

(x+y)(x-y)=(x)²-(y)²

=144

Answer 2

$\large\underline{\sf{Solution-}}$

Given pair of lines are

$\rm :\longmapsto\:x + y = 12 - - (1)$

and

$\rm :\longmapsto\:x - y = 12 - - (2)$

Consider, equation (1),

$\rm :\longmapsto\:x + y = 12$

Substituting 'x = 0' in the given equation, we get

$\rm :\longmapsto\:0 + y = 12$

$\rm :\longmapsto\:y = 12$

Substituting 'y = 0' in the given equation, we get

$\rm :\longmapsto\:x + 0 = 12$

$\rm :\longmapsto\:x = 12$

Hᴇɴᴄᴇ,

➢ Pair of points of the given equation are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 12 \\ \\ \sf 12 & \sf 0 \end{array}} \\ \end{gathered}$

➢ Now draw a graph using the points (0 , 12) & (12 , 0)

➢ See the attachment graph.

Consider, equation (2)

$\rm :\longmapsto\:x - y = 12$

Substituting 'x = 0' in the given equation, we get

$\rm :\longmapsto\:0 - y = 12$

$\rm :\longmapsto\: - y = 12$

$\rm :\longmapsto\: y = - \: 12$

Substituting 'y = 0' in the given equation, we get

$\rm :\longmapsto\:x - 0 = 12$

$\rm :\longmapsto\:x = 12$

Hᴇɴᴄᴇ,

➢ Pair of points of the given equation are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf - 12 \\ \\ \sf 12 & \sf 0 \end{array}} \\ \end{gathered}$

➢ Now draw a graph using the points (0 , - 12) & (12 , 0)

➢ See the attachment graph.

So, from graph we concluded that

The required triangle formed by given lines with y - axis is ABC.

From graph we concluded that,

Area of triangle ABC = 1/2 × 12 × 24 = 144 square units.