The points $A(k,2)$, $B(4,k+1)$ and $C(6,5)$ are collinear. Which of the following is the larger possible value of $k$?

Step-by-step solution

1. For collinearity, slopes $AB$ and $BC$ must be equal.
2. Compute slope $AB$: $m_{AB}=\frac{(k+1)-2}{4-k}=\frac{k-1}{4-k}$.
3. Compute slope $BC$: $m_{BC}=\frac{5-(k+1)}{6-4}=\frac{4-k}{2}$.
4. Set $m_{AB}=m_{BC}$: $\frac{k-1}{4-k}=\frac{4-k}{2}$.
5. Cross‑multiply: $2(k-1)=(4-k{)}^2$.
6. Expand: $2k-2=16-8k+k^2$.
7. Bring all terms to one side: $k^2-10k+18=0$.
8. Solve the quadratic: $k=\frac{10\pm \sqrt{100-72}}{2}=5\pm \sqrt{7}$.
9. The larger value is $k=5+\sqrt{7}$.

Concept tested: Collinearity using equal slopes

Published 23 August 2026 · How this solution was prepared and checked

Related questions

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