Straight LineHardBloom L3
Question
The line $2x + y = 4$ meet $x$-axis at $A$ and $y$-axis at $B$. The perpendicular bisector of $AB$ meets the horizontal line through $(0, - 1)$ at $C$. Let $G$ be the centroid of $\bigtriangleup ABC$. The perpendicular distance from $G$ to $AB$ equals
Options
A.$\sqrt{5}$
B.$\frac{\sqrt{5}}{3}$
C.$2\sqrt{5}$
D.$3\sqrt{5}$
Solution
Sol. [Key Concept: Find the axis-intercepts $A,B$; the perpendicular bisector of $AB$ and the centroid $G$ give the required distance from $G$ to $AB$.]
Slope of $CD$ is $\frac{1}{2} \Rightarrow C \equiv ( - 5, - 1)$
Perpendicular distance from $G$ to $AB = \frac{1}{3}$ (Perpendicular distance from $C$ to $AB$ )
$$= \frac{1}{3}\left( \frac{| - 10 - 1 - 4|}{\sqrt{4} + 1} \right) = \sqrt{5}$$
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