the line segment PQ joining the points P(2,-4) and Q(5,2) is trisected at the points R(3,a) and S(b,0). find the values of a and b.
Answers
Answered by
6
We can solve this by finding the lengths of the segments PR and SQ and PQ. We can find the answers also by slopes.
P (2, -4) R (3, a) S (b,0) Q (5, 2)
PRSQ is a straight line.
Slope of line PR = Slope of PQ
(a+4) / (3-2) = (2+4) / (5-2) = 2
a = -2
Slope of line SQ = Slop of PQ
(2-0) / (5-b) = 2 => 5-b = 1 => b = 4
====================
we can verify if needed the lengths of the line segments :
PR = √(2²+1²) = √5 RS = √5 SQ = √5 PQ = √(6²+3²) = 3√5
P (2, -4) R (3, a) S (b,0) Q (5, 2)
PRSQ is a straight line.
Slope of line PR = Slope of PQ
(a+4) / (3-2) = (2+4) / (5-2) = 2
a = -2
Slope of line SQ = Slop of PQ
(2-0) / (5-b) = 2 => 5-b = 1 => b = 4
====================
we can verify if needed the lengths of the line segments :
PR = √(2²+1²) = √5 RS = √5 SQ = √5 PQ = √(6²+3²) = 3√5
Similar questions