Let the points ABC form a triangle on the cartesian plane, whose area is 20. Let the coordinates of said points be A(8, 6) B(2, 4) and C(x, y) If ∣AC∣=∣BC∣, find the coordinates of point C.


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
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

	$Midpoint of C_{1} C_{2} = M (5, 5)$ $\frac{x + 7}{2} = 5, \frac{y - 1}{2} = 5$ $\Rightarrow x = 3, y = 11$ $C_{2} (x, y) = C_{2} (3, 11)$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum