Solveeit Logo
Login

Question

Question: The length of altitude through A of the triangle ABC, where \(A \equiv ( - 3,0);B \equiv (4, - 1);C ...

The length of altitude through A of the triangle ABC, where A≡(−3,0);B≡(4,−1);C≡(5,2),A \equiv ( - 3,0);B \equiv (4, - 1);C \equiv (5,2), is.

A

210\frac{2}{\sqrt{10}}

B

410\frac{4}{\sqrt{10}}

C

1110\frac{11}{\sqrt{10}}

D

2210\frac{22}{\sqrt{10}}

Answer

2210\frac{22}{\sqrt{10}}

Explanation

Solution

In △ABC\triangle A B C, A≡(−3,0);B≡(4,−1)A \equiv ( - 3,0 ) ; B \equiv ( 4 , - 1 ) and C≡(5,2)C \equiv ( 5,2 )

We know that BC=(5−4)2+(2+1)2B C = \sqrt { ( 5 - 4 ) ^ { 2 } + ( 2 + 1 ) ^ { 2 } } =1+9=10= \sqrt { 1 + 9 } = \sqrt { 10 }

and area of △ABC\triangle A B C

=12[−3(−1−2)+4(2−0)+5(0+1)]=11= \frac { 1 } { 2 } [ - 3 ( - 1 - 2 ) + 4 ( 2 - 0 ) + 5 ( 0 + 1 ) ] = 11

Therefore, altitude AL=2ΔABCBC=2×1110=2210A L = \frac { 2 \Delta A B C } { B C } = \frac { 2 \times 11 } { \sqrt { 10 } } = \frac { 22 } { \sqrt { 10 } } .