JAMB 1997 · UME · Q34

Coordinate geometry

Find the distance between the point $Q(4, 3)$ and the point common to the lines $2x - y = 4$ and $x + y = 2$ .

Worked solution (try it first)

Add the two equations to remove $y$ : $3x = 6$ , so $x = 2$ .
Then $y = 2 - 2 = 0$ , and the lines meet at $(2, 0)$ .
From $(2, 0)$ to $Q(4, 3)$ the changes are $4 - 2 = 2$ and $3 - 0 = 3$ .
By Pythagoras the distance is $\sqrt{2^2 + 3^2} = \sqrt{13}$ , option D.

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