perpendicular bisectors............?

Question

Points A(8, -3), B(-2,-8), C(-4,3) and D(6,8) are the vertices of a quadrilateral.
a)determine the equationof the diagogals
b) is AC the perpendicular bisector of BD? (Explain these to me)
c) is BD the perpendicular bisector of AC?
d) is ABCD a parallelogram of rhombus?

Answer 1

A(8, -3), B(-2, -8), C(-4, 3), D (6, 8)

a) The diagonals are AC and BD.

To find the equation of the line AC, we first find the slope.
(y2 - y1) / (x2 - x1) = m. Through AC, we have (8, -3) (-4, 3). Plug that in for (x1, y1) and (x2, y2).

(3 - (-3)) / (-4 - 8) = m
6/(-12) = m
Therefore, m = -1/2.

The equation of the tangent line is found using the slope formula as well; this time, we're going to use (x1, y1) as (8, -3) and (x2, y2) as (x, y).

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

(y - (-3)) / (x - 8) = (-1/2)
(y + 3) / (x - 8) = -1/2
2(y + 3) = (-1)(x - 8)
y + 3 = (-1/2)(x - 8)
y + 3 = (-1/2)x + 4
y = (-1/2)x + 1

To find the equation of the line BD, we do the same thing with the points (-2, -8) and (6, 8). Using a one-shot method of obtaining the equation of the line through those points, we get

(y - (-8)) / (x - (-2)) = (8 - (-8)) / (6 - (-2))

(y + 8) / (x + 2) = 16/8
(y + 8) / (x + 2) = 2
y + 8 = 2(x + 2)
y + 8 = 2x + 4
y = 2x - 4

b) Our two lines are
AC: y = (-1/2)x + 1
BD: y = 2x - 4

In order for these two lines to be perpendicular, their slopes must be the negative reciprocals of each other, and they are (since -1/2 times 2 = -1).

For AC to be a bisector of BD, the midpoint of AC must satisfy the equation of the line for BD. The midpoint is given by the formula

( (x1 + x2)/2 , (y1 + y2)/2 ). For AC (8, -3) and (-4, 3), we get
( 4/2 , 0/2)
(2, 0)

Does (2, 0) satisfy BD, which is y = 2x - 4?
If x = 2, y = 0, so the answer is yes.
AC is the perpendicular bisector of BD.

c) Is BD the perpendicular bisector of AC? Let's find out.
Take the midpoint of BD: The middle of (-2, -8) and (6, 8).

( (-2 + 6)/2 , (-8 + 8)/2 ) = (4/2, 0/2) = (2, 0)
Does (2, 0) satisfy y = (-1/2)x + 1?
If x = 2, y = (-1/2)(2) + 1 = -1 + 1 = 0, so the answer is yes.
BD is the perpendicular bisector of AC.

d) ABCD is a parallelogram AND a rhombus (because all rhombi are parallelograms). The rhombus has the characteristic that the diagonals not only bisect each other, but do so at 90 degree angles. Therefore, ABCD is a rhombus (which, again, is also a parallogram).

Answer 2

This has already been answered, but there are some errors and I think you might be looking for a bit more explanation:
Gradient of AC is (3-(-3))/(-4-8)
=6/(-12)
= -1/2
Hence equation is
y-(-3) = -(1/2)(x-8) [right so far]
and so
2y+6 = -x +8
2y+x -2 = 0

Using the same method,
equation of BD is
2x-y-4=0

c)As Anilbaks explained, these diagonals are perpendicular as their gradients are -1/2 and 2, whose product is -1.

Also mid-point of AC is
((8-4)/2, (-3+3)/2)
=(2, 0)

Sub this in the equation of BD and find that the point is on the line BDbecause it satisfies the equation:
2*2 - 0 - 4 = 0
i.e. BD passes through the mid-point of AC and is perpendicular to AC, hence it is the perpendicular bisector of AC.

d) I skipped b) -- you do it the same way as c). Now the diagonals of any parallelogram bisect each other, but if also they are perpendicular to each other then the parallelogram is a rhombus. You could check this by using the distance formula on each of the sides and finding that they are all equal (all of them are √125)

Answer 3

Gradient of AC = 3 - (-3)/(-4 - 8) = 6/(-12) = -1/2
Therefore equation of AC using gradient and point A is
y - (-3) = (-1/2)(x - 8) ---> y + 3 = (-1/2)x + 4 ---> y = (-1/2)x + 1
Check that it does it go through C.
when x = -4 ---> y = (-1/2)(-4) + 1 = 2 + 1 = 3 checks.

Gradient of BD = 8 - (-8)/(6 - (-2)) = 16/8 = 2
Therefore equation of BD using gradient and point D is
y - 8 = 2(x - 6) ---> y - 8 = 2x - 12 ---> y = 2x - 4
Check that it does go through B.
when x = -2 ---> y = 2(-2) - 4 = -8 checks

Because gradients of AC and BD multiply to -1 they are perpendicular.

AC intersects BD when 2x - 4 = (-1/2)x +1 ---> x = 2
Using either equation x = 2 ---> y = 0 intersect at (2,0)

Mid point of AC is (8 + (-4))/2, (-3 + 3)/2 = (2,0) so BD does bisect BD.
Mid point of BD is (-2 + 6)/2, (-8 + 8)/2 = (2,0) so AC does bisect BD.

Because the diagonals of the quadrilateral bisect each other at right angles then ABCD is a rhombus.