Graph the line that satisfies each condition.
contains Q(-1,3) and is perpendicular to line AB with A(-2,0) and B(4,3)
How do I do this? lol...
if two lines are perpendicular, then their slopes are the negative inverses of each other. that is, if one's slope is then the other's is
find the slope of the line connecting the last two points. then take it's negative inverse, that will give you the slope of the line you want to make.
then use the point-slope form:
where is the slope of your line, and is a point it passes through
Alright. Then I must have worked the slope wrong... but I was close.
Another (parallel):
contains M(1, -1) and is parallel to FG with F(3,5) and G(-3,-1)
So, I found the slope of F and G which is 4... i think... but I have graphed all the points they've given me, and done the 'rise and run' thing with M, and they are soooooo not parallel. Aren't parallel lines supposed to have the same slope? I didn't find the right slope, did I?
Edit: no wait..... i did do the slope wrong... *fixing slope equation*
Edit 2: The slope is 1... making the two lines parallel.
It would be y = x - 2 wouldn't it?
Okay, on my next problem... they're all making me cry lol.
slope = -4/3, contains K(3, -2)
This is what I got:
m = 1
y + 2 = -4/3(x - 3)
y + 2 = -4/3x - 3
y + 2 = -4/3x + 4
___y = -4/3x + 2
BUT. It tells me to graph it... and how am I supposed to continue the line on from K if it is some whacky slope?! Oi!
yes
the y-intercept is 2. find the x-intercept. then just draw a straight line through those two points, and viola! you're done
Okay, on my next problem... they're all making me cry lol.
slope = -4/3, contains K(3, -2)
This is what I got:
m = 1
y + 2 = -4/3(x - 3)
y + 2 = -4/3x - 3
y + 2 = -4/3x + 4
___y = -4/3x + 2
BUT. It tells me to graph it... and how am I supposed to continue the line on from K if it is some whacky slope?! Oi!
Hello,
you know (or should know ) that the slope of a line is always a fraction and the slope is determined by a right triangle. The legs of the triangle are parallel to the coordinate axes.
With your problem:
That means: Start at K, go 3 units to the right, go 4 units down: There is the a point which belongs to the line. Connect this brand new point with K and you are done.
If you don't like the point K then start at the y-intercept: Go 3 units to the right, 4 units down: There is a point which belongs to the line (and which is known to you: It's good ol' K)