Question

Using a scale of $1 cm$ to $1$ unit for the both the axes, draw the graphs of the following equations:

$6 y = 5 x + 10, y = 5 x - 15$ :

From the graph find:

The area of the triangle between the lines and the $x$ -axis;

Accepted Answer

For the first graph,

we have $6 y = 5 x + 10$

$y = \frac{5 x + 10}{6}$

When $x = - 2, y = \frac{5 (- 2) + 10}{6} = 0$

When $x = 4, y = \frac{5 (4) + 10}{6} = 5$

When $x = - 8, y = \frac{5 (- 8) + 10}{6} = - 5$

$x$ $- 2$ $4$ $- 8$

$y$ $0$ $5$ $- 5$

Thus the ordered pairs of the relation $6 y = 5 x + 10$ are $(- 2, 0), (4, 5)$ and $(- 8, - 5)$ .

Plotting these points on the same graph paper and drawing a line joining them we obtain the graph of the line represented by the equation $6 y = 5 x + 10$ .

For the second graph, $y = 5 x - 15$

When $x = 0, y = 5 (0) - 15 = - 15$

When $x = 1, y = 5 (1) - 15 = - 10$

When $x = 2, y = 5 (2) - 15 = - 5$

$x$ $0$ $1$ $2$

$y$ $- 15$ $- 10$ $- 5$

Thus the ordered pairs of the relation $y = 5 x - 15$ are $(0, 15), (1, - 10)$ and $(2, - 5)$ .

Plotting the points on the same graph paper and drawing a line joining them we obtain the graph of the line represented by the equation $y = 5 x - 15$ .

From the graph, we observe that the line intersect at $(4, 5)$ .

One line intersect the $x$ -axis at $(- 2, 0)$ . and the other line intersect the $x$ axis at $(3, 0)$ .

Take $A = (4, 5), B = (- 2, 0), C = (3, 0)$

From the graph, height of the triangle $= 5$ units

Area of $= \frac{1}{2} \times base \times height$

$= \frac{1}{2} \times 5 \times 5$

$= 12.5 sq . units$ .

O P Malhotra, S K Gupta and, Anubhuti Gangal Solutions for Exercise 3: Exercise

O P Malhotra Mathematics Solutions for Exercise - O P Malhotra, S K Gupta and, Anubhuti Gangal Solutions for Exercise 3: Exercise

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