Question

write an equation for the line perpendicular to the given line that contains p.
P(1,1); y=4x+3

Answer 1

slope of
\[\huge y=\color{blue}mx+b\] is \[\huge \color{blue}m\]
what is the slops of
\[y=4x+3\]?

Answer 2

4

Answer 3

would it be y-1=4(x-1)

Answer 4

ok good
now the "perpendicular" line has a slope that is the negative recrocal
\[\huge -\frac{1}{\color{blue}m}\]

Answer 5

"reciprocal"

Answer 6

what is the slope of the perpendicular line?

Answer 7

i mean the slope would be -1/4

Answer 8

right, so
\[y-1=-\frac{1}{4}(x-1)\] is a start

Answer 9

is that all i need?

Answer 10

What do you think?
The above equation is one form of the desired line.
You could solve this equation for y and end up with the slope-intercept form.

Answer 11

the instructions don't say to put in y intercept it just says to leave it! thank you very much !

Answer 12

satellite knows the following:
The slope (m) x the perpenicular slope(1/m) = -1
its a basic rule in maths.

Answer 13

If you have \[y-1=-\frac{1}{4}(x-1)\] we say this equation is in point-slope form.
Were you to solve for y, you'd get the equivalent equation in slope-intercept form. (You don't "put in y intercept;" rather you IDENTIFY the y-intercept as the constant term on the right: y = mx + b (b is the y-intercept).