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Question
Line PQ is parallel to line RS where points P,Q,R and S have
co-ordinates (2, 4), (3, 6), (3, 1) and (5, k) respectively. Find value of k.
Solution
slope of the line =`(y_2-y_1)/(x_2-x_1)`
P(2, 4), Q(3, 6)
slope of the line PQ = `(6-4)/(3-2) = 2/1 = 2`
R
(3, 1), S (5, k)
slope of the line RS = `(k−1)/(5-3) = (k−1)/2`
But line PQ || line RS
∴ slope of line PQ = slope of line RS
∴ `2 = (k−1)/2`
∴ 4 = k - 1
∴ 4 + 1 = k
∴ k = 5
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