Simple Calculus

**fvaras89** · August 16th 2009, 10:37 PM

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

**pickslides** · August 16th 2009, 10:53 PM

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

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

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

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

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

Now use (4,-1) to find c and you will have your answer....

**fvaras89** · August 16th 2009, 10:58 PM

Thank you so much!!

Is it alright to have the final equation as

y+1 = -5/4 (x-4)?

**pickslides** · August 16th 2009, 11:03 PM

Its ok, depends what form you are asked to provide your answer in.

Your answer is in the form $y-y_1 = m(x-x_1)$ which is fine.

If you need to find the y-intercept which is common you would need to find c as follows

$y=\frac{-5}{4}x+c$

$-1=\frac{-5}{4}\times 4+c$

$-1=-5+c$

$c=4 \Rightarrow y=\frac{-5}{4}x+4$

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