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I think this is an extremely simple problem, but my answer is different from the answer sheet. I wonder whether I miss some concept. My answer is 24, but the answer sheet is 12.

Determine the value of $k$ so that the points P(-8,10), Q(2,-4), and R(22,k) are collinear.

learning
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2 Answers2

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There are many ways to do this. For things to be collinear, their slopes have to be the same. The slope of $PQ=\frac{-7}{5}$. Then the slope of $QR $ must equal the slope of $PQ$.$\frac{k+4}{20}=\frac{-7}{5}$. Therefore, $k=-32$.

AfronPie
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  • Read this if you need more help http://math.stackexchange.com/questions/405966/if-i-have-three-points-is-there-an-easy-way-to-tell-if-they-are-collinear. – AfronPie Oct 04 '16 at 03:47
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Note that $P$, $Q$, and $R$ are colinear if and only if the vectors $\vec{PQ}$ and $\vec{PR}$ are parallel. But two vectors are parallel if and only if their cross product is the zero vector. Hence $P$, $Q$, and $R$ are colinear if and only if $$ \vec 0=\vec{PQ}\times\vec{PR}=\left(10,\,-14,\,0\right)\times \left(30,\,k - 10,\,0\right)=\left(0,\,0,\,10 \, k + 320\right) $$ Since $10\,k+320=0$ if and only if $k=-32$, it follows that $P$, $Q$, and $R$ are colinear if and only if $k=-32$.

Brian Fitzpatrick
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