https://gyazo.com/f4b5386358a6510a10b219672a2f0cd0 a) I found their slopes and they were perpendicular. b) 3x+2y-11=0 c) r = (5,-2)+t(2,3) Did I do it correctly?
I'm a little rusty on this but I believe to find the line with vector d1 passing through (5,-2) you would apply the vector to point (5,-2) and find the equation between (5+2,-2+3) and (5,-2) which is slightly different to yours I believe a and c are good though
@angle sorry to bother but could you double check me?
I think, in terms of turning in an answer, you can add specific numbers for (a) because the question asks to "Show" so you say something like: d1 has a slope of _____ and d2 has a slope of ____ and since they are flipped opposites of each other, this means the vectors are perpendicular.
Yeah, I gave an explanation for that.
For b tho: I had direction vector: (2,3) and point (5,-2) 2x+3y+C = 0 2(5)+3(-2)+C = 0 C = -4 2x+3y-4=0 I think that's the right answer. Am I correct?
So that would change my answer for C as well
|dw:1527861082437:dw|
so this line has a slope = 3/2 and goes through points (5, -2) and (7, 1)
I think it's becomes a different equation than the one you came up with
But I don't see how I can be wrong. :\
I'm not sure what formula you're using, but I can try to show you in "point slope form" y = m(x - h) + k where m = slope and (h, k) a point in the line y = (3/2)(x - 5) - 2 [plugged in values] 2y = 3(x - 5) - 4 [multiplied everything by 2] 2y = 3x - 15 - 4 [distributed the 3] 2y = 3x - 19 [added like terms] 3x - 2y - 19 = 0 [subtracted 2y from both sides] so 3x - 2y - 19 = 0
so you were off by a just little in your first attempt :)
Okay, so I was trying with slope intercept form and I got this: y=3/2x+b -2=3/2(5)+b b = -2-3/2(5) b = -19/2 y=3/2x-19/2 ==> Multiply each side by 2 2y = 3x-19 2y-3x+19 = 0 I don't get what I did wrong.
it right you can multiply through by -1 to get the same answer that I just got :)
Aha I see. So both equations can work. Tysm!!
mhm ^_^
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