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September 16, 2014

September 16, 2014

Posted by **Anonymous** on Monday, September 10, 2012 at 2:11pm.

Please help!

a) find the equatoin of line 1 which passes through (1,3) and (9,7)

I get y=3+1/2(x+1)or y=1/2x+2.5

b) the line 2 is perpedicular to line 1 and passes through 5,0. find the equation

for this one i get y=-2x+10

c) find where line 1 and line 2 intersect each other. use this to compute the distance between the point (5,0) and line 1

I get (-5,0) for the intersection, which doesnt make sense to me?

Thanks

- precalc -
**Steve**, Monday, September 10, 2012 at 3:08pmline 1:

(y-3)/(x-1) = (7-3)/(9-1)

y-3 = 1/2 (x-1)

y = 1/2 x + 5/2

you are correct

line 2:

(y-0) = -2 (x-5)

y = -2x + 10

you are correct

1/2 x + 5/2 = -2x + 10

5/2 x = 15/2

x = 3 so y=4

The lines intersect at (3,4)

d^2 = (3-5)^2 + (4-0)^2

d^2 = 4+16 = 20

d = 2√5

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