Question

What is the slope of the line that goes through (5, -1) and is perpendicular to a line with slope 2/5.

Answer 1

-5/2

Explanation

Step 1

find the slope of the line :

when 2 lines are perpendicular the product of their slopes equals -1,so

$\begin{gathered} if\text{ line 1}\perp\text{ line 2} \\ then \\ slope_1*slope_2=-1 \end{gathered}$

then, let

$\begin{gathered} slope_1=given=\frac{2}{5} \\ slope_2=slope_2 \\ hence \\ \frac{2}{5}*slope_2=-1 \\ to\text{ solve, multiply both sides by 5/2} \\ \frac{2}{5}slope_2*\frac{5}{2}=-1*\frac{5}{2} \\ slope_2=-\frac{5}{2} \end{gathered}$

therefore, the slope of the line we are looking for is -5/2

Step 2

now, use the point-slope formula to find the equation of the line, it says

$\begin{gathered} y-y_1=m(x-x_1) \\ where\text{ m is the slope } \\ (x_1,y_1)\text{ is a point from the line} \end{gathered}$

a)let

$\begin{gathered} slope=slope_2=-\frac{5}{2} \\ point=(5,-1) \end{gathered}$

b) replace and solve for y to find the equation

$\begin{gathered} y-y_{1}=m(x-x_{1}) \\ y-(-1)=-\frac{5}{2}(x-5) \\ y+1=-\frac{5}{2}x+\frac{25}{2} \\ subtract\text{ 1 in both sides} \\ y+1-1=-\frac{5}{2}x+\frac{25}{2}-1 \\ y=-\frac{5}{2}x+\frac{23}{2} \end{gathered}$

therefore, the equation of the line is

$y=-\frac{5}{2}x+\frac{23}{2}$

and the slope is

$-\frac{5}{2}$

the slope of the line we are looking for is -5/2