Question

Find an equation for the line containing the points $\left(-2,5\right)$and $(4,-1)$. What is the distance between these points? What is their midpoint?

Short answer

The equation of the line is $x+y=3$.

The distance between these two points is $6\sqrt{2}$.

Their midpoint is$\left(1,2\right)$.

Step-by-step solution

Step 1. Given Information

We are given the points $\left(-2,5\right)$and $\left(4,-1\right)$. We need to find equation for the line containing these two points, the distance between these points and their midpoint.

Step 2. Finding equation of the line

Let the two points be

$$\begin{gathered} (x_1,y_1)=(-2,5) \\ (x_2,y_2)=(4,-1) \end{gathered}$$

Equation of a line is given by

$$\begin{gathered} y-y_1=\left(\frac{x-x_1}{x_2-x_1}\right)\left(y_2-y_1\right) \\ \Rightarrow y-5=\left(\frac{x-\left(-2\right)}{4-\left(-2\right)}\right)\left(-1-5\right) \\ \Rightarrow y-5=\left(\frac{x+2}{4+2}\right)\left(-6\right) \\ \Rightarrow y-5=\left(\frac{x+2}{6}\right)\left(-6\right) \\ \Rightarrow y-5=-(x+2) \\ \Rightarrow x+y=5-2 \\ \Rightarrow x+y=3 \end{gathered}$$

Step 3. Finding the distance

Distance between two points $\left(x_1,y_1\right)$and $\left(x_2,y_2\right)$is given by

$$\begin{gathered} \sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2} \\ =\sqrt{{\left(4-\left(-2\right)\right)}^2+{\left(-1-5\right)}^2} \\ =\sqrt{{\left(4+2\right)}^2+{\left(-6\right)}^2} \\ =\sqrt{{\left(6\right)}^2+36} \\ =\sqrt{36+36} \\ =\sqrt{72} \\ =2\sqrt{6} \end{gathered}$$

Step 4. Finding the midpoint

The midpoint of two points is given by $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$.

$$\begin{gathered} \therefore \left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right) \\ =\left(\frac{-2+4}{2},\frac{5-1}{2}\right) \\ =\left(\frac{2}{2},\frac{4}{2}\right) \\ =\left(1,2\right) \end{gathered}$$