Question

Find the equation of the line passing through $(-3,5)$ and perpendicular to the line through the points $(2,5)$ and $(-3,6)$.

Answer 1

Step1: Calculation of the equation of line passing through $(2,5)$ and $(-3,6)$.

The two-point form of the equation of line passing through $(x_1,y_1)$ and $(x_2,y_2)$ is given by the formula $\left(y-y_1\right)=\frac{(y_2-y_1)}{(x_2-x_1)}(x-x_1)$.

For the line passing through $(2,5)$ and $(-3,6)$, $x_1=2,y_1=5,x_2=-3,y_2=6$.

Substitute these points in equation $\left(y-y_1\right)=\frac{(y_2-y_1)}{(x_2-x_1)}(x-x_1)$ and simplify.

$$\begin{gathered} y-5=\frac{(6-5)}{(-3-2)}(x-2) \\ \Rightarrow y-5=\frac{1}{(-5)}(x-2) \\ \Rightarrow y-5=-\frac{1}{5}(x-2) \\ \Rightarrow y-5=-\frac{1}{5}x+\frac{2}{5} \\ \Rightarrow y=-\frac{1}{5}x+\frac{2}{5}+5(adding5tobothsides) \\ \Rightarrow y=-\frac{1}{5}x+\frac{2+25}{5} \\ \therefore y=-\frac{1}{5}x+\frac{27}{5}-equation\left(1\right) \end{gathered}$$

Step2: Calculation of the slope of line.

The slope-intercept from of the equation of a line is given by the formula $y=mx+c$, where $m$ is the slope and $c$ is the $Y$-intercept of the line.

Comparing equation (1) with the equation $y=mx+c$ we get the slope of the line given in equation (1) as $m=-\frac{1}{5}$.

The slope of any line perpendicular to the line given in equation (1) is the negative reciprocal of $m=-\frac{1}{5}$.

Let the slope of the perpendicular line be $m\text{'}$.

$$\begin{gathered} m\text{'}=-\frac{1}{m} \\ \Rightarrow m\text{'}=-\frac{1}{\left({\displaystyle \raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}\right)} \\ \Rightarrow m\text{'}=-1(-5) \\ \therefore m\text{'}=5 \end{gathered}$$

Step3: Calculation of the equation of line passing through the point $(-3,5)$ and perpendicular to $y=-\frac{1}{5}x+\frac{27}{5}$.

The point-slope form of the equation of a line passing through $(x_1,y_1)$ and having slope $m$ is given by the formula $\left(y-y_1\right)=m(x-x_1)$.

Thus, the equation of a line passing through $(-3,5)$ and having slope $5$ is:

$$\begin{gathered} \left(y-5\right)=5(x-(-3\left)\right) \\ \Rightarrow (y-5)=5(x+3) \\ \Rightarrow y-5=5x+15 \\ \Rightarrow y-5x-5-15=0(subtracting5x+15frombothsides) \\ \Rightarrow y-5x-20=0 \\ \therefore 5x-y+20=0(multiplyingbothsidesby-1) \end{gathered}$$

Hence, the equation of the line passing through the point $(-3,5)$ and perpendicular to the line through the points $(2,5)$ and $(-3,6)$ is $5x-y+20=0$.