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Question
The slope of a line joining P(6,k) and Q(1−3k,3) is 12. Find: |k|
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Solution
Let P(6,k)=(x1,y1) and Q(1−3k,3)=(x2,y2)
∴ Slope of PQ=y2−y1x2−x1
⇒3−k1−3k−6=12
6−2k=−5−3k
11=−k
k=−11∴|k|=11
∴ Slope of PQ=y2−y1x2−x1
⇒3−k1−3k−6=12
6−2k=−5−3k
11=−k
k=−11
∴|k|=11
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