line 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
I;m not sure if i am doing this right...I keep getting a negative intersection point, which doesn't seem possible.
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
1 answer
5 answers