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The points A(1, 2), B(3, k) and C(7, 14) lie on one straight line. Find k.

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

The points $A(1, 2)$ , $B(3, k)$ and $C(7, 14)$ lie on one straight line. Find $k$ .

Need a reminder? The key ideaShow

The area of the triangle with vertices $(x_i, y_i)$ is $\tfrac{1}{2}\left|\det\right|$ of the matrix with rows $(x_i, y_i, 1)$ . Three points are collinear exactly when this determinant is $0$ . A matrix multiplies every area by $|\det T|$ .

The worked solution

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