Question

draw a graph of x minus y + 1 is equal to zero 3 X + 2 Y - 12 is equal to zero calculate the area bounded by these lines and x axis​

Answer 1

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

Given pair of linear equations are

$\rm :\longmapsto\:x - y + 1 = 0 - - - (1)$

and

$\rm :\longmapsto\:3x + 2y - 12 = 0 - - - (2)$

Consider equation (1)

$\rm :\longmapsto\:x - y + 1 = 0$

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

$\rm :\longmapsto\:0 - y + 1 = 0$

$\rm :\longmapsto\: - y + 1 = 0$

$\rm :\longmapsto\: - y= - 1$

$\bf\implies \:y = 1$

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

$\rm :\longmapsto\:x - 0 + 1 = 0$

$\rm :\longmapsto\:x + 1 = 0$

$\bf\implies \:x = - 1$

Hᴇɴᴄᴇ,

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

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

➢ Now draw a graph using the points (0 , 1) & (- 1 , 0)

➢ See the attachment graph.

Consider equation (2)

$\rm :\longmapsto\:3x + 2y - 12 = 0$

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

$\rm :\longmapsto\:3(0) + 2y - 12 = 0$

$\rm :\longmapsto \: 2y - 12 = 0$

$\rm :\longmapsto \: 2y = 12$

$\bf\implies \:y = 6$

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

$\rm :\longmapsto\:3x + 2(0) - 12 = 0$

$\rm :\longmapsto\:3x - 12 = 0$

$\rm :\longmapsto\:3x = 12$

$\bf\implies \:x = 4$

Hᴇɴᴄᴇ,

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

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

➢ Now draw a graph using the points (0 , 6) & (4 , 0)

➢ See the attachment graph.

Now, from graph we concluded that, the required triangle formed by the given two lines with x - axis is ABC having vertices

Coordinates of A (2, 3)

Coordinates of B (- 1, 0)

Coordinates of C (4, 0)

So, Base of triangle ABC = 5 units

Height of triangle ABC = 3 units

So, area of triangle ABC is

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

$\rm \:  =  \: \dfrac{15}{2}$

$\rm \:  =  \:7.5 \: square \: units$