# Simple Calculus

• Aug 16th 2009, 09:37 PM
fvaras89
Simple Calculus
Find an equation for the line through (4,-1) and perpendicular to the line through (-2,3) and (3,7).

Please provide all steps as im very confused about what to do (Crying)
• Aug 16th 2009, 09:53 PM
pickslides
Quote:

Originally Posted by fvaras89
Find an equation for the line through (4,-1) and perpendicular to the line through (-2,3) and (3,7).

Please provide all steps as im very confused about what to do (Crying)

first of all find the gradient for the points (-2,3) and (3,7).

$\displaystyle m= \frac{y_2-y_1}{x_2-x_1} = \frac{7-3}{3-(-2)} = \frac{4}{5}$

perpendicualr gradient is $\displaystyle \frac{-1}{m} = \frac{-1}{\frac{4}{5}} = \frac{-5}{4}$

So the line $\displaystyle y=mx+c$ for the perpendicular becomes $\displaystyle y=\frac{-5}{4}x+c$

Now use (4,-1) to find c and you will have your answer....
• Aug 16th 2009, 09:58 PM
fvaras89
Thank you so much!!

Is it alright to have the final equation as

y+1 = -5/4 (x-4)?
• Aug 16th 2009, 10:03 PM
pickslides
Your answer is in the form $\displaystyle y-y_1 = m(x-x_1)$ which is fine.
$\displaystyle y=\frac{-5}{4}x+c$
$\displaystyle -1=\frac{-5}{4}\times 4+c$
$\displaystyle -1=-5+c$
$\displaystyle c=4 \Rightarrow y=\frac{-5}{4}x+4$