Question

Write the equation of the line perpendicular to the line x - 5y = -10 and passing through the point (2,5).
I really honestly just can't even..

Answer 1

So one way of writing normal lines \(y=mx+b\) where \(m\) is the slope and \(b\) is the y-intercept. So we know that the first line is actually:
\[\eqalign{
&x-5y=-10 \\
&x=5y-10 \\
&x+10=5y \\
&\frac{x+10}{5}=y \\
&0.2x+2=y\\
&y=0.2x+2
}\]

Answer 2

Ok, how do I figure out the answer?

Answer 3

So you know that the line has a slope of \(0.2\)
If the slope of the original line was \(m\), the slope of the perpendicular line is \(-\frac{1}{m}\). This means that the slope of the perpendicular line is: \(-\frac{1}{0.2}=-5\).
So so far the equation of the perpendicular line is:
\[y=-5x+b\]
We need to find \(b\). Lets sub in a point such as the one we are given \((2,5)\):
\[\eqalign{
&5=-5(2)+b \\
&5=-10+b \\
&5+10=b \\
&15=b \\
}\]
So therefore the equation is:
\[y=-5x+15\]
Or a version which looks nicer:
\[y=15-5x\]

Answer 4

Can you help me with a couple more?

Answer 5

Ehh...Maybe...Yeah I could. Where?