Question

Draw the graphs of the following equations on the same graph sheet:

x + 1 = 0, y + 1 = 0, x + y = 5

. Also, find the area enclosed between these lines.

Accepted Answer

The given equations are $x + 1 = 0, y + 1 = 0, x + y = 5$ .

Here, the equation $x + 1 = 0$ represents a line parallel to Y-axis and at a distance of $1$ unit to the left of it.

Similarly, the equation $y + 1 = 0$ represents a line parallel to X-axis and at a distance of $1$ unit below it.

Now we get some points to plot $x + y = 5$ .

For $x = - 2$ , $y = 5 - x = 5 - (- 2) = 5 + 2 = 7$ . So one point on the line is $(- 2, 7)$ .

Similarly for $y = - 2$ , $x = 5 - y = 5 - (- 2) = 7$ . So another point on the given line is $(7, - 2)$ .

Now, the table for the points to be plotted on the graph is given below.

X-values Y-values

$- 2$ $7$

$7$ $- 2$

The graph is drawn below.

Here for each of the two axes, the distance between two consecutive whole number points is $1$ unit.

The intersection points of the three lines are $(- 1, 6), (- 1, - 1) & (6, - 1)$ . These points form a triangle.

Area enclosed between the lines is the area of triangle.

Area of triangle $= \frac{1}{2} \times base \times height = \frac{1}{2} \times (6 - \{- 1\}) \times (6 - \{- 1\}) = \frac{1}{2} \times 7 \times 7 = \frac{49}{2} = 24.5 square units$

Therefore the required area is $24.5 square units .$

A K Upadhyay and S Upadhyay Solutions for Chapter: Linear Equations in Two Variable, Exercise 2: EXERCISE

A K Upadhyay Mathematics Solutions for Exercise - A K Upadhyay and S Upadhyay Solutions for Chapter: Linear Equations in Two Variable, Exercise 2: EXERCISE

Questions from A K Upadhyay and S Upadhyay Solutions for Chapter: Linear Equations in Two Variable, Exercise 2: EXERCISE with Hints & Solutions

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