Using a scale of 1 cm to 1 unit for both the axes, draw the graphs of the following equations: 6y = 5x + 10, y = 5x - 15. From the graph, find

(i) the coordinates of the point where the two lines intersect.

(ii) the area of the triangle between the lines and the x-axis.

(i) Given,

⇒ 6y = 5x + 10

⇒ y = $\dfrac{5x + 10}{6}$ ………(1)

When x = 1, y = $\dfrac{5 \times 1 + 10}{6} = \dfrac{15}{6}$ = 2.5

x = -2, y = $\dfrac{5 \times -2 + 10}{6} = \dfrac{-10 + 10}{6} = \dfrac{0}{6}$ = 0,

x = 4, y = $\dfrac{5 \times 4 + 10}{6} = \dfrac{20 + 10}{6} = \dfrac{30}{6}$ = 5.

Table of values of equation (1) :

Steps of construction :

1. Plot the points (1, 2.5), (-2, 0) and (4, 5) on graph.

2. Join the points.

Given,

y = 5x - 15

When x = 2.5, y = 5 × 2.5 - 15 = 12.5 - 15 = -2.5,

x = 3, y = 5 × 3 - 15 = 15 - 15 = 0,

x = 4, y = 5 × 4 - 15 = 20 - 15 = 5.

Table of values of equation (2) :

Steps of construction :

1. Plot the points (2.5, -2.5), (3, 0) and (4, 5) on graph.

2. Join the points.

From graph,

The lines intersect at point P(4, 5).

Hence, x = 4, y = 5.

(ii) From graph,

Triangle = PQR.

Draw a line PJ, from P perpendicular to x-axis.

PJ = 5 units

QR = 5 units

Area of triangle = $\dfrac{1}{2}$ × base × height

= $\dfrac{1}{2}$ × QR × PJ

= $\dfrac{1}{2}$ × 5 × 5

= $\dfrac{25}{2}$

= 12.5 sq. units.

Hence, area = 12.5 sq. units.