Answer:
Step-by-step explanation:
The vertices of the square given are P(-4, 0), Q(4, 3), R(7, -5) and, S(-1, -18)
For this diagonal to be right angle the slope of the diagonal must be m1=-1/m2
So let find the slope of diagonal 1
The two points are P and R
P(-4, 0), R(7, -5)
Slope is given as
m1=∆y/∆x
m1=(y2-y1)/(x2-x1)
m1=-5-0/7--4
m1=-5/7+4
m1=-5/11
Slope of the second diagonal
Which is Q and S
Q(4, 3), S(-1, -18)
m2=∆y/∆x
m2=(y2-y1)/(x2-x1)
m2=(-18-3)/(-1-4)
m2=-21/-5
m2=21/5
So, slope of diagonal 1 is not equal to slope two
This shows that the diagonal of the square are not diagonal.
But the diagonal of a square should be perpendicular, this shows that this is not a square, let prove that with distance between two points
Given two points
(x1,y1) and (x2,y2)
Distance between the two points is
D=√(y2-y1)²+(x2-x1)²
For line PQ
P(-4, 0), Q(4, 3)
PQ=√(3-0)²+(4--4)²
PQ=√(3)²+(4+4)²
PQ=√9+8²
PQ=√9+64
PQ=√73
Also let fine RS
R(7, -5) and, S(-1, -18)
RS=√(-18--5)+(-1-7)
RS=√(-18+5)²+(-1-7)²
RS=√(-13)²+(-8)²
RS=√169+64
RS=√233
Since RS is not equal to PQ then this is not a square, a square is suppose to have equal sides
But I suspect one of the vertices is wrong, vertices S it should have been (-1,-8) and not (-1,-18)
So using S(-1,-8)
Let apply this to the slope
Q(4, 3), S(-1, -8)
m2=∆y/∆x
m2=(y2-y1)/(x2-x1)
m2=(-8-3)/(-1-4)
m2=-11/-5
m2=11/5
Now,
Let find the negative reciprocal of m2
Reciprocal of m2 is 5/11
Then negative of it is -5/11
Which is equal to m1
Then, the square diagonal is perpendicular
