P is the point (a, a-2) and Q is the point (4-3a, -a).
A.) Find the gradient of the line PQ.
B.) Find the gradient of a line perpendicular to PQ.
C.) Given that the distance PQ is 10√5, find the two possible values of a
Please show your work.
Answers
Given : P is the point (a, a-2) and Q is the point (4-3a, -a).
distance PQ is 10√5,
To Find : A.) Find the gradient of the line PQ.
B.) Find the gradient of a line perpendicular to PQ.
C.) Given that the distance PQ is 10√5, find the two possible values of a
Solution:
P ( a , a - 2) , Q ( 4 - 3a , - a)
Gradient = Slope = (-a - (a - 2) ) / ( 4 - 3a - a)
= (2 - 2a) / (4 - 4a)
= (2 - 2a)/ 2(2 - 2a)
= 1/2
Gradient PQ = 1/2
gradient of a line perpendicular to PQ. = m
=> m * 1/2 = - 1
=> m = - 2
gradient of a line perpendicular to PQ. = - 2
distance PQ is 10√5
P ( a , a - 2) , Q ( 4 - 3a , - a)
=> (4 - 3a - a)² + ( -a - (a - 2))² = (10√5)²
=> (4 - 4a)² + (2-2a)² = 500
=> 16(1 - a)² + 4(1-a)² = 500
=> 20(1 - a)² = 500
=> (1 - a)² = 25
=> 1 - a = ± 5
=> a = 6 , a = - 4
Possible values of a are -4 & 6
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