Question

(iii) Draw the graph representing the equations x - y = 1 and 2x + 3y = 12 on the same
graph paper. Find the area of the triangles formed by these lines, the X-axis and
the Y-axis.​

Answer 1

Given :-

• First equation of line

$\boxed{ \green{ \bf \: x - y = 1}}$

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

$\begin{gathered}:\implies\:\:\sf{0\: - \:y\:=\:1} \\ \end{gathered}$

$\begin{gathered}:\implies\:\:\bf{y\:= \: - \:1} \\ \end{gathered}$

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

$\begin{gathered}:\implies\:\:\sf{x\: - \:0\:=\:1} \\ \end{gathered}$

$\begin{gathered}:\implies\:\:\bf{x\:=\:1} \\ \end{gathered}$

Hᴇɴᴄᴇ,

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

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf - 1 \sf & \sf \\ \\ \sf 1 & \sf 0 \end{array}} \\ \end{gathered}$

Now,

• Second equation of line is

$\boxed{ \pink{ \bf \: 2x + 3y = 12}}$

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

$\begin{gathered}:\implies\:\:\sf{0\:+3\:y\:=\:12} \\ \end{gathered}$

$\begin{gathered}:\implies\:\:\bf{y\:=\:4} \\ \end{gathered}$

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

$\begin{gathered}:\implies\:\:\sf{2x\:+\:0\:=\:12} \\ \end{gathered}$

$\begin{gathered}:\implies\:\:\bf{x\:=\:6} \\ \end{gathered}$

❸ Substituting 'x = 3' in the given equation, we get

$\begin{gathered}:\implies\:\:\sf{2 \times 3\:+3\:y\:=\:12} \\ \end{gathered}$

$\begin{gathered}:\implies\:\:\sf{6\:+3\:y\:=\:12} \\ \end{gathered}$

$\begin{gathered}:\implies\:\:\sf{3\:y\:=\:6} \\ \end{gathered}$

$\begin{gathered}:\implies\:\:\bf{y\:=\:2} \\ \end{gathered}$

Hᴇɴᴄᴇ,

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

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 4 \\ \\ \sf 3 & \sf 2 \\ \\ \sf 6 & \sf 0 \end{array}} \\ \end{gathered}$

➢ See the attachment graph.

Now area of triangle bounded by lines with x - axis is

$\rm \: \rightarrow \: \dfrac{1}{2} \times 6 \times 2$

$\rm \: \rightarrow \: 6 \: square \: units$

Now, area of triangle bounded by lines with y - axis is

$\rm \: \rightarrow \: \dfrac{1}{2} \times 5 \times 3$

$\rm \: \rightarrow \: \dfrac{15}{2} \: square \: units.$