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 and a light ray eliminating from point A goes along the direction of internal bisector of the angle A, which meets the mirror and BC at E, D respectively. If sum of the areas of `triangle ACE` and `triangle ABE` is `K` sq units then`(2K)/5-6` is
A
17 sq. units
B
18 sq. units
C
`(50)/(3)` sq. units
D
20 sq. units
Text Solution
Verified by Experts
The correct Answer is:
C
`(AD)/(DE) =(BD)/(DC) =(AB)/(AC)` `:.` Point `D ((1)/(3),(1)/(3))` and `E(-2,0)`. Area of `DeltaADC +` Area of `DeltaBDE` `=(25)/(3) +(25)/(3)` `=(50)/(3)` sq. units
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A straight line parallel to the base BC of the triangle ABC intersects AB and AC at the points D and E respectively. If the area of the Delta ABE be 36 sq. cm, then the area of the Delta ACD is
CENGAGE-COORDINATE SYSTEM-Multiple Correct Answers Type
