ABCD is a quadrilateral with angle ABC a right angle. The point D lies on the perpendicular bisector of AB. The coordinates of A and B are (7, 2) and (2, 5) respectively. The equation of line AD is y = 4x − 26. find the area of quadrilateral ABCD

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1. Equation of the perpendicular bisector of line segment AB: y=5x/3 - 4
Coordinates of point D = (66/7, 82/7)
Gradient of BC = 5/3
Coordinates of point C = (8,15)
These are the right answers but how to find the area of quadrilateral? Apparently the answer is 629/14 square units

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2. There are several ways to do this:

1. If you join AC, you have a right-angled triangle ABC. Find the length of AB and BC and use
area = (1/2)base x height for its area.
Find angle D by using the slopes of AD and CD,
find the lengths of CD and AD , then area of triangle ACD = (1/2)(CD)(AD)sin D

2. You could use Heron's formula to find the area of triangle ACD, add on the area of the right-angled triangle

3. The simplest and quickest way to find the area of any convex polygon is this:
list the coordinates of the quadrilateral clockwise starting with any point in a column.
repeat the first point you started with.
Area = (1/2)(sum of the diagonal downproducts - sum of the diagonal upproducts)

for yours:
7 2
66/7 82/7
8 15
2 5
7 2

Area = (1/2)[ (7(82/7) + 15(66/7) + 40 + 4) - (2(66/7) + 8(82/7) + 30 + 35)
= (1/2)[1872/7 - 1243/7]
= (1/2)(629/7)
= 629/14

The last method works for any polygon.
Listing the points in a clockwise rotation will result in the negative of the above, so just take the absolute value if you ignore the rotation.
The important thing is to list the points in sequence, and to repeat the first one you started with.

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3. For the last method, I meant to say :

" list the coordinates of the quadrilateral counter-clockwise starting with any point in a column. "

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4. I am not sure how you got 40+4 and 30+35
btw thx for your help

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5. never mind i got it- it's just like matching

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