Answer:
Question 2:
No. The point (1,-6) is not on line L.
Step-by-step explanation:
A straight line L passes through points (a,12) and (4,a)
L is perpendicular to line whose equation is 2x - 3y + 4 = 0
Putting the equation in the straight line equation form:
3y = 2x + 4
y = 2/3x +4/3
Since slope(m) of the perpendicular line is 2/3, line L has a slope of -1 ÷ 2/3 = -3/2
Finding this slope using points given on L:
(a - 12) ÷ (4 - a) = -3/2
2a - 24 = -12 + 3a
a = -12
Taking a point on L (4,a) which is (4,-12) and another point (1,-6);
slope(m) of L = -3/2 and should be equal to (-6 - -12) ÷ (1 - 4) = 6/-3 = -2
Clearly -3/2 ≠ -2
Since a point (1,6) does not give a slope(m) of -3/2 with a point on L; then the point is not on L.
