Maths M.C.Q

False
Solution:
The coordinates of p0oints are (0, 2), (1, 3), (2, 4), (3, –5) and (4, 6).
Given equation is x0 - y + 2 = 0
At point (0, 2), 0 – 2 + 2 = 0 ⇒ 0=0, it satisfies.
At point (1, 3), 1 - 3 + 2 = 3-3 = 0 ⇒ 0 = 0, it satisfies.
At point (2, 4), 2 - 4 + 2 = 4 - 4 = 0 ⇒ 0 = 0, it satisfies.
At point (3, -5), 3 – (- 5) + 2 = 3 + 5 + 2 = 10 ≠ 0, it does not satisfy.
At point (4, 6), 4 - 6 + 2 - 6 - 6 = 0 ⇒ 0 = 0, it satisfies.
Hence, point (3, – 5) does not satisfy the equation.

False
Solution:
Since, the graph of a linear equation in two variables always represent a line.

True
Solution:
If the given points (-1, 1) and (- 3, 3) lie on the linear equation x + y = 0, then both points will satisfy the equation.
So, at point (-1, 1), we put x = -1, and y = 1 in LHS of the given equation, we get
LHS = x + y = -1 + 1 = 0 = RHS
Again, at point (-3, 3) put x = -3 and y = 3 in LHS of the given equation, we get,
LHS = x + y= -3 + 3 = 0 = RHS
Hence, (-1, 1) and (-3, 3) both satisfy the given linear equation x + y = 0.

False

Solution:

If we put x = 0 and y = 7 in LHS of the given equation, we get,

LHS = (0) + 2 (7)= 0 + 14 = 14 ≠ 7 = RHS

Hence, (0, 7) does not lie on the line x + 2y = 7.

True

Solution:

If we put x = 0 and y = 3 in LHS of the given equation, we find,

LHS = 3 × 0 + 4 × 3 = 0 + 12 = 12 = RHS

Hence, (0, 3) lies on the linear equation 3x + 4y = 12.

False
Solution:
Since, every point on the graph of the linear equation represents a solution.

True
Solution:
given graph is a line parallel to y-axis at a distance 3 units to the right of the origin.
Hence, it represents a linear equation x = 3.