Question

on a coordinate plane, triangle a b c is shown. point a is at (negative 1, 1), point b is at (3, 2), and points c is at (negative 1, negative 1)if line segment bc is considered the base of triangle abc, what is the corresponding height of the triangle?

Answer 1

Answer:

The corresponding height of the triangle is $1.2$

Step-by-step explanation:

We have$\Delta ABC$

Here $AD$ is the height

We need to draw perpendicular from $A$ to $BC$

The distance from the point $(m, n)$ to the line $A x+B y+C=0$ is given by:

$d=\frac{|A m+B n+C|}{\sqrt{A^{2}+B^{2}}}$

First let us calculate equation of $BC$

$y-y_1=m(x-x_1)\\\\y-2=\frac{2--1}{3--1} (x-3)\\\\y-2=\frac{3}{4} (x-3)\\\\4y-8=3x-9\\\\3x-4y=1$

Now distance from $(-1,1)$ to $3x-4y-1=0$ is

$d=\frac{|A m+B n+C|}{\sqrt{A^{2}+B^{2}}}\\\\=\frac{|3 \times1+-4\times-1-1|}{\sqrt{3^{2}+(-4)^{2}}}\\\\=\frac{6}{5}\\\\=1.2$