Coordinate Geometry Question 101
Question: The length of altitude through A of the triangle ABC, where $ A\equiv (-3,0);B\equiv (4,-1);C\equiv (5,2), $ is
[Karnataka CET 2001]
Options:
A) $ \frac{2}{\sqrt{10}} $
B) $ \frac{4}{\sqrt{10}} $
C) $ \frac{11}{\sqrt{10}} $
D) $ \frac{22}{\sqrt{10}} $
Show Answer
Answer:
Correct Answer: D
Solution:
In $ \Delta ABC $ , $ A\equiv (-3,0);B\equiv (4,-1) $ and $ C\equiv (5,2) $ We know that $ BC=\sqrt{{{(5-4)}^{2}}+{{(2+1)}^{2}}} $
$ =\sqrt{1+9}=\sqrt{10} $ and area of $ \Delta ABC $
$ =\frac{1}{2}[-3(-1-2)+4(2-0)+5(0+1)]=11 $ Therefore, altitude $ AL=\frac{2\Delta ABC}{BC}=\frac{2\times 11}{\sqrt{10}}=\frac{22}{\sqrt{10}} $ .
BETA
