Find the coordinates of the intersection of the diagonals of parallelogram GHJK with vertices G (-2, 3), H (4, 4), J (2, -1), and K(-4, - 2).

The point of intersection of both diagonals is (0,1)

Step-by-step explanation:

for the parallelogram, we have the diagonals GJ and HK

Firstly, we need to get the equation of the lines that join these two points

For GJ

the general equation is;

y = mx + b

m is slope and b is y-intercept

We have the points

(-2,3) and (2,-1)

The slope m is as follows;

m = (y2-y1)/(x2-x1)

m = (-1-3)/(2-(-2) = -4/4 = -1

So we have;

y = -x + b

To get b, we substitute any of the two points

Let us use (-2,3)

x = -2 and y = 3

so we have;

3 = -1(-2) + b

3 = 2 + b

b = 3-2 = 1

so we have the equator GJ as y = -x + 1

We repeat same for HK

(4,4) and (-4,-2)

m = (-2-4)/(-4-4) = -6/-8 = 3/4

so we have;

y = 3x/4 + b

use (4,4) to get b

4 = 3/4(4) + b

4 = 3 + b

b = 1

so we have the equate;

y = 3x/4 + 1

multiply through by 4

4y = 3x + 4

To get the point of intersection, we have to solve the equations of the diagonals simultaneously

the equations are;

4y = 3x + 4

y = -x + 1

Multiply equation ii by 4 and i by 1

4y = 3x + 4

4y = -4x + 4

Subtract ii from i

0 = 7x

x = 0

To get y, use any of the equations;

y = -x + 1

y = 0 + 1

y = 1

So the point of intersection of both diagonals are;

(0,1)

the answer is two. hope this !