Asked by Sal
The points A,B,C,D have coordinates (3,3) (8,0) (-1,1) (-6,4) respectively.
Find the coordinates of the point of intersection of the diagonals AC and BD?
Help plZ!!!
Find the coordinates of the point of intersection of the diagonals AC and BD?
Help plZ!!!
Answers
Answered by
Reiny
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
<b>x - 2y = -3</b>
Now you find the equation for BD
then solve the two equations.
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
<b>x - 2y = -3</b>
Now you find the equation for BD
then solve the two equations.
Answered by
Sal
I get the gradient part reiny, but how do you use one of the points to get the equation?
Answered by
Sal
The gradient/slope for BD is -2/7
Answered by
Sal
Ah wait i got the equation for BD as
y=-2/7 x + 16/7
is that right?
y=-2/7 x + 16/7
is that right?
Answered by
Reiny
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
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
Answered by
Reiny
Yes, your second equation is correct