Question

Draw the graphs of equation x-y = -1 and 3x +2y = 12.
Determine the coordinates of vertices of the triangle formed by these lines and the x-axis and shade the triangular region.

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Answer 1

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

Given pair of linear equations are

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

and

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

Consider First Equation

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

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

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

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

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

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

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

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

Hᴇɴᴄᴇ,

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

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

Consider Second Equation

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

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

$\rm :\longmapsto\:3 \times0 + 2y = 12$

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

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

$\rm :\longmapsto\:y = 6$

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

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

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

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

$\rm :\longmapsto\:x = 4$

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

➢ See the attachment graph.

Now, from graph we concluded that the required area bounded by the given lines with x - axis is ABC having coordinates as

Coordinates of A (2, 3)

Coordinates of B (4, 0)

Coordinates of C (- 1, 0)

Answer 2

Answer:

38 94

Step-by-step explanation: