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

The corresponding height of the triangle is
`1.2`

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{\sqrt{A^(2)+B^(2)}}`

First let us calculate equation of
`BC`

`y-y_1=m(x-x_1)\\\\y-2=(2--1)/(3--1) (x-3)\\\\y-2=(3)/(4) (x-3)\\\\4y-8=3x-9\\\\3x-4y=1`

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

`d=\fracA m+B n+C{\sqrt{A^(2)+B^(2)}}\\\\=\frac3 *1+-4*-1-1{\sqrt{3^(2)+(-4)^(2)}}\\\\=(6)/(5)\\\\=1.2`