Answer:
Explanation:
The diagonals of a quadrilateral divide it into two congruent triangles. In this case, the diagonals of quadrilateral ABCD with vertices A (-4,9), B (3,9), C(2,3), D(-5,3) are AC and BD.
To find the intersection point of the two diagonals, we can use the equations of the two lines.
The equation of line AC can be represented as y = mx + b, where m is the slope of the line and b is the y-intercept.
m = (y2-y1)/(x2-x1)
m = (3-9)/(2-(-4)) = -6/6 = -1
b = y1 - mx1
b = 9 - (-1)(-4) = 9+4 = 13
The equation of the line is y = -x + 13
Similarly, the equation of line BD can be represented as y = nx + c, where n is the slope of the line and c is the y-intercept.
n = (y4-y3)/(x4-x3) = (3-9)/(-5-3) = -6/8 = -3/4
c = y3 - nx3
c = 3 - (-3/4)(-5) = 3 + 15/4 = 27/4
The equation of the line is y = -3/4 x + 27/4
Now we can find the point of intersection (x,y) by solving the following system of equations
-x + 13 = -3/4 x + 27/4
Multiply the equation 1 by 4
-4x + 52 = -3x + 27
Add 3x to both sides
-x + 52 = 27
Subtract 27 from both sides
-x = -25
x = 25
Now we can substitute this value of x in any of the equation to find the value of y
y = -x + 13
y = -25 + 13
y = -12
So, the coordinates of the intersection of the diagonals of quadrilateral ABCD are (25,-12).