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A triangle ABC has vertices $A (5, 1), B (- 1, - 7),$ and $C (1, 4)$ respectively. L be the line mirror passing through C and parallel to AB. A light ray emanating from point A goes along the direction of the internal bisector of angle A, which meets the mirror and BC at E and D, respectively. Then the sum of the areas of $ΔACE$ and $ΔABC$ is

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Detailed Solution

$\frac{AD}{DE} = \frac{BD}{DC} = \frac{AB}{AC} = 2$ $∴ D \equiv (\frac{1}{3}, \frac{1}{3})$

and $E \equiv (- 2, 0)$

Area of $ΔACE$ + Area of $ΔABC$

$= \frac{25}{2} + 25 = 37 .5 sq . units$

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