Question

5. If a linear equation has solutions (-2, 2), (0,0) and (2,-2), then its is of the form (a) y- x = 0 (b) x +y= 0 (c) -2x +y= 0 (d) –x+2y= 0​

Answer 1

Linear equations

Method 1. We simply substitute the solution in each of the options

When x = - 2, y = 2

• (a) y - x = - 2 - 2 = - 4 ≠ 0
• (b) x + y = - 2 + 2 = 0
• (c) - 2x + y = - 2 (- 2) + 2 = 4 + 2 = 6 ≠ 0
• (d) - x + 2y = - 2 + 2 (- 2) = - 2 - 4 = - 6 ≠ 0

We see that only option (b) is satisfied by the solution (- 2, 2).

Now we check for other solutions as well.

When x = 0, y = 0, x + y = 0 + 0 = 0

When x = 2, y = - 2, x + y = 2 - 2 = 0

So option (c) is correct.

Method 2. Geometrical approach

The equation of the straight line possibly through the points (- 2, 2) and (0, 0) is given by

(y - 2)/(2 - 0) = (x + 2)/(- 2 - 0)

or, (y - 2)/2 = (x + 2)/(- 2)

or, y - 2 = - x - 2

or, x + y = 0

When x = 2, y = - 2,

x + y = 2 - 2 = 0

Option (c) is correct.