SOLUTION: Find the point D(x,y) such that the points A(-3,1), B(4,0), C(0,-3) and D are the corners of a square. Justify your answer. ****** Start of solution: ****** Distance formul

Algebra ->  Parallelograms -> SOLUTION: Find the point D(x,y) such that the points A(-3,1), B(4,0), C(0,-3) and D are the corners of a square. Justify your answer. ****** Start of solution: ****** Distance formul      Log On


   

Question 773915: Find the point D(x,y) such that the points A(-3,1), B(4,0), C(0,-3) and D are the corners of a square. Justify your answer.
****** Start of solution: ******
Distance formula:
+d+=+sqrt%28+%28Xsub1+-+Xsub0%29%5E2+%2B+%28Ysub1+-+Ysub0%29%5E2++%29++
So side AC of the square:
= sqrt( (0 - (-3))^2 + (-3-1)^2 )
= sqrt( 9 + 16 )
= 5
I don't know how to continue. . .
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
This can be done without using Distance Formula, since the four points are vertices of a square. (You CAN use the distance formula if you want.)

Segment AD // segment CB
Line AD has slope %280-%28-3%29%29%2F%284-0%29=3%2F4
Using point A(-3,1) and slope 3/4 in point-slope form equation,
y-1=%283%2F4%29%28x-%28-3%29%29
y-1=%283%2F4%29%28x%2B3%29
y=%283%2F4%29%28x%2B3%29%2B1
y=%283%2F4%29x%2B9%2F4%2B4%2F4
Line AD______highlight%28y=%283%2F4%29x%2B13%2F4%29

Segment BD // segment CA
Line BD has slope, since is perpendicular to line AD, of -%284%2F3%29.
Using point B(4,0) and slope -(4/3) in point-slope form equation,
y-0=-%284%2F3%29%28x-4%29
Line BD_____highlight%28y=-%284%2F3%29x%2B16%2F3%29

Point D(?,?) is the intersection of the two lines, highlight%28y=%283%2F4%29x%2B13%2F4%29 and highlight%28y=-%284%2F3%29x%2B16%2F3%29.


------remaining process would be solve the system.
------easiest way would be clear the fractions and use the equations in standard form..