JAMB 2012 · UTME · Q33

The distance between the point $(4, 3)$ and the intersection of $y = 2x + 4$ and $y = 7 - x$ is

Worked solution (try it first)

At the intersection both $y$ -values are equal: $2x + 4 = 7 - x$ , so $3x = 3$ and $x = 1$ .
Then $y = 7 - 1 = 6$ , so the lines meet at $(1, 6)$ .
From $(1, 6)$ to $(4, 3)$ the changes are 3 and $-3$ , so the distance is $\sqrt{9 + 9} = \sqrt{18} = 3\sqrt2$ , option B.