Question

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Answer 1

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.