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Question Number 178374 by depressiveshrek last updated on 16/Oct/22

L e t t h e p o i n t s A B C f o r m a t r i a n g l e o n t h e

c a r t e s i a n p l a n e, w h o s e a r e a i s 20 . L e t t h e c o o r d i n a t e s

o f s a i d p o i n t s b e A (8, 6) B (2, 4) a n d C (x, y)

I f ∣ A C ∣=∣ B C ∣, f i n d t h e c o o r d i n a t e s o f p o i n t C .

Answered by Ar Brandon last updated on 16/Oct/22

Let M be the midpoint of B C . Then

M has coordinates (\frac{8 + 2}{2}, \frac{6 + 4}{2}) = M (5, 5)

If ∣ A C ∣=∣ B C ∣ then Δ_{A B C} is an isosceles triangle

∣ A M ∣= \sqrt{{(8 - 5)}^{2} + {(6 - 5)}^{2}} = \sqrt{10}

∣ C M ∣= \sqrt{{(x - 5)}^{2} + {(y - 5)}^{2}}

∣ A C ∣=∣ B C ∣

\Rightarrow \sqrt{{(x - 8)}^{2} + {(y - 6)}^{2}} = \sqrt{{(x - 2)}^{2} + {(y - 4)}^{2}}

\Rightarrow - 6 (2 x - 10) = 2 (2 y - 10) \Rightarrow y - 5 = - 3 (x - 5)

\Rightarrow∣ C M ∣= \sqrt{{(x - 5)}^{2} + 9 {(x - 5)}^{2}} = \sqrt{10} ∣ x - 5 ∣

Area of triangle = 20

\Rightarrow \frac{1}{2} ∣ A M ∣ \cdot ∣ C M ∣= 20 \Rightarrow \frac{1}{2} \sqrt{10} \times \sqrt{10} ∣ x - 5 ∣= 10

\Rightarrow x = 7 \Rightarrow y - 5 = - 3 (7 - 5) \Rightarrow y = - 1

Hence coordinates of C are C (7, - 1)

Commented by depressiveshrek last updated on 16/Oct/22

C o r r e c t, b u t t h e r e i s a l s o a n o t h e r p o i n t o f C t h a t

h e l p s t o f o r m t h e t r i a n g l e .

Commented by Ar Brandon last updated on 16/Oct/22

Commented by Ar Brandon last updated on 16/Oct/22

The second point is the reflection of C_{1} on y_{A B}

Commented by Ar Brandon last updated on 16/Oct/22

Midpoint of C_{1} C_{2} = M (5, 5)

\frac{x + 7}{2} = 5, \frac{y - 1}{2} = 5

\Rightarrow x = 3, y = 11

C_{2} (x, y) = C_{2} (3, 11)