question_answer

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]

A) \[\frac{2}{\sqrt{10}}\]

B) \[\frac{4}{\sqrt{10}}\]

C) \[\frac{11}{\sqrt{10}}\]

D) \[\frac{22}{\sqrt{10}}\]

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 \[AL=\frac{2\,\Delta ABC}{BC}=\frac{2\times 11}{\sqrt{10}}=\frac{22}{\sqrt{10}}\].