Given-point-A-7-26-and-B-12-12-find-all-points-P-such-that-AP-BP-and-APB-90-

Question Number 119007 by bramlexs22 last updated on 21/Oct/20

G i v e n p o i n t A = (7, 26) a n d B = (12, 12), f i n d

a l l p o i n t s P s u c h t h a t ∣ A P ∣ = ∣ B P ∣ a n d ∠ A P B = 90 ° .

Answered by bemath last updated on 21/Oct/20

N o t e t h a t t h e t r i a n g l e A B P i s a n i s o s c e l e s r i g h t

t r i a n g l e w i t h ∣ A P ∣ = ∣ B P ∣ . L e t Q b e t h e m i d p o i n t

o f s e g m e n t A B . T h e n Q = (\frac{19}{2}, 19), w h e r e

∣ Q A ∣ = ∣ Q B ∣ = ∣ Q P ∣ a n d Q A ⊥ Q B

T h u s \vec{Q} \vec{A} = (\frac{5}{2}, - 7) a n d \vec{Q P} = (7, \frac{5}{2})

o r (- 7, - \frac{5}{2}) . L e t O = (0, 0) b e t h e o r i g i n, i t f o l l o w s t h a t

O \vec{P} = O \vec{Q} + Q \vec{P} = (\frac{19}{2}, 19) \pm (7, \frac{5}{2}) .

C o n s e q u e n t l y P = (\frac{33}{2}, \frac{43}{2}) o r (\frac{5}{2}, \frac{33}{2})

Answered by 1549442205PVT last updated on 21/Oct/20

\vec{AB} = (5, - 14)

Midpoint of AB is M (\frac{19}{2}, 19), AP = BP

\Rightarrow P (x, y) lying on the midline d of AB, so

\vec{MP} = (\frac{19}{2} - x, 19 - y) perpendicular to

\vec{AB} \Leftrightarrow \vec{AB} . \vec{MP} = 0 \Leftrightarrow 5 (\frac{19}{2} - x) - 14 (19 - y) = 0

\Leftrightarrow 5 x - 14 y + 218.5 = 0 \Rightarrow P (x, \frac{5 x + 218.5}{14})

\vec{AP} = (x - 7, \frac{5 x - 145.5}{14}), \vec{BP} = (x - 12, \frac{5 x + 50.5}{14})

\hat{APB} = 90 ° \Leftrightarrow \vec{AP} ⊥ \vec{BP} \Leftrightarrow \vec{AP} . \vec{BP} = 0

\Leftrightarrow (x - 7) (x - 12) + (\frac{5 x - 145.5}{14}) (\frac{5 x + 50.5}{14}) = 0

x^{2} - 19 x + 84 + \frac{{25 x}^{2} - 475 x - 7347.25}{196} = 0

{221 x}^{2} - 4199 x + 9116.75 = 0

x_{1} = 16.5, x_{2} = 2.5 . We get two the pointP

P_{1} (16.5, \frac{43}{2}), P_{2} (2.5, \frac{33}{2})