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.
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slope of line PQ =slope of line RS
y2-y1/x2- x1 =y2-y1/x2- x1
6-4/3-2 = k-1/5-3
2 /1 = k-1/2
2×2 =k-1
4= k-1
4+1=k
Hence, k =5
y2-y1/x2- x1 =y2-y1/x2- x1
6-4/3-2 = k-1/5-3
2 /1 = k-1/2
2×2 =k-1
4= k-1
4+1=k
Hence, k =5
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