Question

Find the equations of the sides of a triangle whose vertices are at A (-1, 8), B (4, -2) and C (-5, -3).

Answer 1

Hint: Every side of a triangle has two endpoints or vertices. A line is drawn through these points to make a side. Here the points are given in the question we just have to find the equation of the line through these points using two points form of an equation

Complete step-by-step answer:
Two points form of a linear equation is $y-y_1=\left({\displaystyle \frac{y_2-y_1}{x_2-x_1}}\right)\left(x-x_1\right)$ where the first point is $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is the second point.
We are given a triangle with vertices A (-1, 8), B (4, -2) and C (-5, -3).

We have to find the equations of the sides of the above triangle ABC.
Equation when two points are given is $y-y_1=\left({\displaystyle \frac{y_2-y_1}{x_2-x_1}}\right)\left(x-x_1\right)$
Equation of side AB= $y-y_1=\left({\displaystyle \frac{y_2-y_1}{x_2-x_1}}\right)\left(x-x_1\right)$
Where $$\begin{gathered} \left(x_1,y_1\right)=A(-1,8)=(-1,8) \\ \left(x_2,y_2\right)=B(4,-2)=(4,-2) \end{gathered}$$
  $$\begin{gathered} y-8=\left({\displaystyle \frac{-2-8}{4-\left(-1\right)}}\right)\left(x-\left(-1\right)\right) \\ y-8=\left({\displaystyle \frac{-10}{4+1}}\right)\left(x+1\right) \\ y-8=\left({\displaystyle \frac{-10}{5}}\right)\left(x+1\right) \\ y-8=\left(-2\right)\left(x+1\right) \\ y-8=-2x-2 \\ 2x+y-8+2=0 \\ 2x+y-6=0 \end{gathered}$$
Equation of side AB is $2x+y-6=0$
Equation of side BC= $y-y_1=\left({\displaystyle \frac{y_2-y_1}{x_2-x_1}}\right)\left(x-x_1\right)$
Where $$\begin{gathered} \left(x_1,y_1\right)=B(4,-2)=(4,-2) \\ \left(x_2,y_2\right)=C(-5,-3)=(-5,-3) \end{gathered}$$
  $$\begin{gathered} y-\left(-2\right)=\left({\displaystyle \frac{-3-\left(-2\right)}{-5-4}}\right)\left(x-4\right) \\ y+2=\left({\displaystyle \frac{-3+2}{-9}}\right)\left(x-4\right) \\ y+2=\left({\displaystyle \frac{-1}{-9}}\right)\left(x-4\right) \\ y+2=\left({\displaystyle \frac{1}{9}}\right)\left(x-4\right) \\ 9\left(y+2\right)=x-4 \\ 9y+18=x-4 \\ x-9y-4-18=0 \\ x-9y-22=0 \end{gathered}$$
Equation of side BC is $x-9y-22=0$
Equation of side CA= $y-y_1=\left({\displaystyle \frac{y_2-y_1}{x_2-x_1}}\right)\left(x-x_1\right)$
Where $$\begin{gathered} \left(x_1,y_1\right)=C(-5,-3)=(-5,-3) \\ \left(x_2,y_2\right)=A(-1,8)=(-1,8) \end{gathered}$$
   $$\begin{gathered} y-\left(-3\right)=\left({\displaystyle \frac{8-\left(-3\right)}{-1-\left(-5\right)}}\right)\left(x-\left(-5\right)\right) \\ y+3=\left({\displaystyle \frac{8+3}{-1+5}}\right)\left(x+5\right) \\ y+3=\left({\displaystyle \frac{11}{4}}\right)\left(x+5\right) \\ y+3=\left({\displaystyle \frac{11}{4}}\right)\left(x+5\right) \\ 4\left(y+3\right)=11\left(x+5\right) \\ 4y+12=11x+55 \\ 11x-4y+55-12=0 \\ 11x-4y+43=0 \end{gathered}$$
Equation of side CA is $11x-4y+43=0$
Equations of sides AB, BC, CA are $2x+y-6=0$ , $$x-9y-22=0$$, \[11x - 4y + 43 = 0 $ respectively.

Note: To form a line we at least need two points. A line is defined as a line of points that extends infinitely in two directions. It has one dimension, length. Points that are on the same line are called collinear points. A line is written with an arrowhead.