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. Alight 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 △ACE and ΔABC is $$________$$ .

Question

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. Alight 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 △ACE and ΔABC is $$________$$ .

Infinity Learn · Accepted Answer

$$ADDE=BDDC=ABAC$$∴ D≡$$13$$,$$13$$ and $$E=($$−$$2$$,$$0)$$ Area of △$$ACE+$$ Area of △$$ABC=2$$× Area of △$$ACE=2$$×$$192$$ sq.units

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$\begin{array}{l} \frac{AD}{DE} = \frac{BD}{DC} = \frac{AB}{AC} \\ ∴ D \equiv (\frac{1}{3}, \frac{1}{3}) and E = (- 2, 0) \end{array}$

$\begin{array}{l} Area of △ ACE + Area of △ ABC \\ = 2 \times Area of △ ACE \\ = 2 \times \frac{19}{2} sq . units \end{array}$

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Cartesian plane

Remember concepts with our Masterclasses.

ADDE=BDDC=ABAC∴ D≡13,13 and E=(−2,0) Area of △ACE+ Area of △ABC=2× Area of △ACE=2×192 sq.units

Ready to Test Your Skills?

free study material

$\begin{array}{l} \frac{AD}{DE} = \frac{BD}{DC} = \frac{AB}{AC} \\ ∴ D \equiv (\frac{1}{3}, \frac{1}{3}) and E = (- 2, 0) \end{array}$

$\begin{array}{l} Area of △ ACE + Area of △ ABC \\ = 2 \times Area of △ ACE \\ = 2 \times \frac{19}{2} sq . units \end{array}$