If a linear equation has solutions (–2, 2), (0, 0) and (2, – 2), then it is of the form
(A) y – x = 0
(B) x + y = 0
(C) –2x + y = 0
(D) –x + 2y = 0
Answer:
(B) x + y = 0
Solution:
(A) Put the given points in y – x = 0
Putting (–2, 2)
LHS = 2 – (–2)
= 2 + 2
The given equation does not satisfy this point, so there is no need to check for other points.
(B) Now put the given points in x + y = 0
Putting (–2, 2)
LHS = –2 + 2 = 0 (RHS)
Putting (0, 0)
LHS = 0 + 0 = 0 (RHS)
Putting (2, –2)
LHS = 2 – 2 = 0 (RHS)
The given equation is satisfying all the points
(C) Put the given points in –2x + y = 0
Putting (–2, 2)
LHS = (–2) (–2) + (2)