144k views
0 votes
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?

User Ssimeonov
by
6.1k points

1 Answer

3 votes

Answer:

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

on a coordinate plane, triangle a b c is shown. point a is at (negative 1, 1), point-example-1
User BenW
by
7.1k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.