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Question 1127556: The coordinates of A are(-3,2) And the coordinates of C are(5,6) .The midpoint of AC is M and the perpendicular bisector of AC cuts the x-axis at B.
Given that ABCD is a square,find the coordinates of D and the length of AD.
Answer by greenestamps(13214)   (Show Source):
You can put this solution on YOUR website!
A(-3,2), C(5,6) --> midpoint M is (1,4) [x coordinate is average of -3 and 5; y coordinate is average of 2 and 6]
To find the slope of AC, I strongly recommend drawing a picture instead of plugging numbers into the slope formula. It's very easy to get wrong numbers in the wrong places in the formula; it's easy to get the right slope using a picture.
From A to C is 8 to the right and up 4; that makes the slope 4/8 = 1/2. The slope of the perpendicular is then -2.
Again use the picture to find where the perpendicular bisector crosses the x-axis, instead of plugging numbers into a formula. The slope is -2; the midpoint M is at (1,4), 4 units above the x-axis. If the slope is -2 and you need to go down 4 units to get to the x-axis, you need to move 2 to the right; 2 to the right from 1 is x=3.
So point B is at (3,0).
The problem says it is given that ABCD is a square; but it is easy to show that AB and BC are two sides of a square: A to B is 6 right and 2 down; B to C is 2 right and 6 up. The picture shows those two segments are perpendicular, and that they have the same length of sqrt(40), or 2*sqrt(10).
And to find the coordinates of D, we again use the picture. From B to C is 2 right and 6 up; if ABCD is a square, then A to D is also 2 right and 6 up. 2 right and 6 up from A(-3,2) is (-1,8).
ANSWERS: Point D is (-1,8); the length of AD is the length of either AB or BC, which is 2*sqrt(10).
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