Question

draw the graph of the equation y=-1,y=3,4x-y=5.also find the area of the quadrilateral formed by the lines and Y-axis.​

Answer 1

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

Given lines are

$\rm :\longmapsto\:4x - y = 5$

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

$\rm :\longmapsto\:y = 3$

Consider line (1),

$\rm :\longmapsto\:4x - y = 5$

On substituting y = 3, we get

$\rm :\longmapsto\:4x - 3 = 5$

$\rm :\longmapsto\:4x = 8$

$\rm :\longmapsto\:x = 2$

On substituting x = 0, we get

$\rm :\longmapsto\:0 - y = 5$

$\rm :\longmapsto\:y = - 5$

On substituting x = 1, we get

$\rm :\longmapsto\:4 - y = 5$

$\rm :\longmapsto\: - y = 5 - 4$

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

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

Hᴇɴᴄᴇ,

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

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf - 5 \\ \\ \sf 1 & \sf - 1 \\ \\ \sf 2 & \sf 3 \end{array}} \\ \end{gathered}$

➢ Now draw a graph using the points.

➢ See the attachment graph.

Consider Line (2),

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

Its a line parallel to x - axis which passes through the point (0, - 1).

➢ Now draw a graph using the points.

➢ See the attachment graph.

Consider Line (3),

$\rm :\longmapsto\:y = 3$

Its a line parallel to x - axis which passes through the point (0, 3).

➢ Now draw a graph using the points.

➢ See the attachment graph.

Now, from graph we concluded that the figure thus formed by these three lines with y - axis is trapezium.

The coordinates of trapezium ABCD are

• Coordinates of A be (1, - 1)

• Coordinates of B (2, 3)

• Coordinates of C (0, 3)

• Coordinates of D (0, - 1).

[ Please see the attachment ].

From graph we concluded that

• BC = 2 units

• CD = 4 units

• DA = 1 units

Thus,

Area of trapezium = 1/2 ( Sum of || sides ) × distance between them

So,

Area of trapezium = 1/2 × ( AD + BC ) × CD

Area of trapezium = 1/2 × ( 2 + 1 ) × 4 = 3 × 2 = 6 sq. units.