Answer:
- B lies on the perpendicular bisector of AC
- D = (10, 0)
- 1/3
Step-by-step explanation:
Given A(0, 10), B(2, 16), C(8, 14), you want to show ∆ABC is isosceles, and you want point D on the x-axis that makes ABCD a kite. Finally, you want the ratio of areas of ∆ABC to ∆ACD.
Perpendicular bisector
If ∆ABC is isosceles, point B will lie on the perpendicular bisector of AC. The equation for that line can be found using ...
(∆x, ∆y) = C -A = (8, 14) -(0, 10) = (8, 4)
The midpoint of AC is ...
M = (A +C)/2 = ((0, 10) + (8, 14))/2 = (8, 24)/2 = (4, 12)
Then the equation of the line through (4, 12) can be written as ...
(∆x)(x -4) +(∆y)(y -12) = 0
8(x -4) +4(y -12) = 0
8x -32 +4y -48 = 0 . . . . . next, we write this in standard form
2x +y = 20 . . . . [line 'a']
Isosceles
∆ABC is isosceles if point B(2, 16) lies on this line:
2x +y = 20
2(2) +(16) = 20 . . . . . . true; B lies on line 'a'
Point D
The x-coordinate of point D is the point on the perpendicular bisector where y = 0:
2x + 0 = 20
x = 10
Point D is (10, 0).
Area ratio
We can look at the vectors BM and MD to determine the ratio of areas of ∆ABC and ∆ADC.
BM = B -M = (2, 16) -(4, 12) = (-2, 4)
MD = M -D = (4, 12) -(10, 0) = (-6, 12)
We see that BM = 1/3 × MD. Since the area is proportional to the height of the triangle, this means the ratio of areas is ...
Area ABC / Area ADC = 1/3
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Additional comment
We could find the lengths AB and CB and show they are equal.
We could equate the distances from D to A and to C, solving for the value of x to make it true. We prefer not to use the distance formula for these purposes if there is a simpler way.
If you have a line with direction = (∆x, ∆y), the perpendicular line through a point (a, b) can be written as ...
(∆x)(x -a) +(∆y)(y -b) = 0
as we have done above.
