Posted by Sal on Wednesday, January 30, 2013 at 7:51pm.
You will have to find the equation of both diagonals
I will do AC , using the points (3,3) and (-1,1)
slope of AC = (3-1)/(3+1) = 2/4 = 1/2
then again using (3,3)
y-3 = (1/2)(x-3)
2y - 6 = x-3
x - 2y = -3
Now you find the equation for BD
then solve the two equations.
I get the gradient part reiny, but how do you use one of the points to get the equation?
The gradient/slope for BD is -2/7
Ah wait i got the equation for BD as
y=-2/7 x + 16/7
is that right?
Maybe you use y = mx + b
then y = (1/2)x + b
sub in (3,3)
3 = (1/2)(3) + b
b = 3 - 3/2
b = 3/2
y = (1/2)x + 3/2
suppose I multiply each term by 2 to get
2y = x + 3
re-arrange for x - 2y = -3 , the same as I had before, but much easier and faster
The method of y = mx + b seems to be the one taught most often these days, but personally, I would hardly ever use it if my slope is a fraction.
If the slope is an integer, then it makes sense to use
y = mx + b
Yes, your second equation is correct
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