Straight Line Question 76

Question: The vertices of a triangle are (2, 1), (5, 2) and (4, 4). The lengths of the perpendicular from these vertices on the opposite sides are [IIT 1962]

Options:

A) $ \frac{7}{\sqrt{5}},\frac{7}{\sqrt{13}},\frac{7}{\sqrt{6}} $

B) $ \frac{7}{\sqrt{6}},\frac{7}{\sqrt{8}},\frac{7}{\sqrt{10}} $

C) $ \frac{7}{\sqrt{5}},\frac{7}{\sqrt{8}},\frac{7}{\sqrt{15}} $

D) $ \frac{7}{\sqrt{5}},\frac{7}{\sqrt{13}},\frac{7}{\sqrt{10}} $

Show Answer

Answer:

Correct Answer: D

Solution:

  • $ L _{12}\equiv x-3y+1=0 $ $ L _{23}\equiv 2x+y-12=0 $ $ L _{13}\equiv 3x-2y-4=0 $ Therefore, the required distances are $ D_3=| \frac{4-3\times 4+1}{\sqrt{10}} |=\frac{7}{\sqrt{10}} $ $ D_1=| \frac{4+1-12}{\sqrt{5}} |=\frac{7}{\sqrt{5}} $ $ D_2=| \frac{3\times 5-2\times 2-4}{\sqrt{9+4}} |=\frac{7}{\sqrt{13}} $ .