Question 11826: hi im having trouble on this problem. "graph the line perpendicular to the graph of 3x-2y=24 that intersects it at its x-intercepts." what i did was get the equation into the yintercept form i think its called like that. y=-12+3/2x. the x-intercept is 3/2 right? and i plot the -12 on the graph right? now i dont know what to do. thank you
Answer by longjonsilver(2297)   (Show Source):
You can put this solution on YOUR website! no, the straight line in "y-intercept" form or whatever you want to call it is:
y = (3/2)x - 12... always think of it as y = mx + c ie x term first.
so, the gradient is 3/2 and the y-intercept is -12.
If you are going to plot this properly, then we need 2 points to join. We now have (0, -12).
The second point we can find easily is the x-intercept, when y=0...
3x-2y=24
3x = 24
x = 24/3
--> x=8
--> so the point is (8,0)
This allows you to plot the first equation. How about plotting the second, the equation that is perpendicular?
well, it will fit the line y=mx+c again. We need its gradient and y-intercept.
Its gradient is -2/3... 2 lines perpendicular to each other will have their gradients m and n where mn=-1. This is ALWAYS true
(eg if one gradient is 3, the other will be -1/3 etc)
so, we have y = -(2/3)x + c
To find c, we need to know a (x,y) pair, which we do, because we know the line goes through the point (8,0), so 0 = -(2/3)(8) + c
-16/3 + c = 0
c = 16/3
--> equation is y = -(2/3)x + 16/3
now to find another point on this line, to draw the straight line... pick the easiest one you can, as ever... so when x=0, y = 16/3, so now you can plot the line.
jon.
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