Math Question and Answers


Question and Answers Forum

All Questions Topic List
Coordinate Geometry Questions
Previous in All Question Next in All Question
Previous in Coordinate Geometry Next in Coordinate Geometry
Question Number 47513 by ajfour last updated on 11/Nov/18

Commented by ajfour last updated on 11/Nov/18

$F i n d c o o r d i n a t e s o f a l l p o i n t s$ $t h a t h a v e s a m e ⊥ d i s t a n c e f r o m$ $f r o m t h e t h r e e l i n e s .$ $F o r e a c h s u c h p o i n t t h i s ⊥$ $d i s t a n c e m a y d i f f e r .$
	Answered by mr W last updated on 11/Nov/18

	$A (c, d)$ $B (0, 0)$ $C (a, b)$ $B C = u = \sqrt{a^{2} + b^{2}}$ $C A = v = \sqrt{{(a - c)}^{2} + {(b - d)}^{2}}$ $A B = w = \sqrt{c^{2} + d^{2}}$ $s = (u + v + w) / 2$ $e q n . o f A B :$ $\frac{y}{x} = \frac{d}{c}$ $\Rightarrow d x - c y = 0$ $e q n . o f B C :$ $\frac{y}{x} = \frac{b}{a}$ $\Rightarrow b x - a y = 0$ $I n c i r c l e w i t h i n c e n t e r I :$ $r_{I} = \frac{Δ}{s} = \sqrt{\frac{(s - u) (s - v) (s - w)}{s}}$ $d_{⊥} f r o m I t o A B = r_{I} :$ $\frac{{d x}_{I} - {c y}_{I}}{\sqrt{c^{2} + d^{2}}} = s i g n (a d - b c) r_{I}$ $\Rightarrow {d x}_{I} - {c y}_{I} = s i g n (a d - b c) r_{I} \sqrt{c^{2} + d^{2}}$ $d_{⊥} f r o m I t o B C = r_{I} :$ $\frac{{b x}_{I} - {a y}_{I}}{\sqrt{a^{2} + b^{2}}} = s i g n (b c - a d) r_{I}$ $\Rightarrow {b x}_{I} - {a y}_{I} = - s i g n (a d - b c) r_{I} \sqrt{a^{2} + b^{2}}$ $(a d - b c) x_{I} = s i g n (a d - b c) r_{I} (a \sqrt{c^{2} + d^{2}} + c \sqrt{a^{2} + b^{2}})$ $\Rightarrow x_{I} = \frac{(a \sqrt{c^{2} + d^{2}} + c \sqrt{a^{2} + b^{2}}) r_{I}}{∣ a d - b c ∣}$ $(a d - b c) y_{I} = s i g n (a d - b c) r_{I} (b \sqrt{c^{2} + d^{2}} + d \sqrt{a^{2} + b^{2}})$ $\Rightarrow y_{I} = \frac{(b \sqrt{c^{2} + d^{2}} + d \sqrt{a^{2} + b^{2}}) r_{I}}{∣ a d - b c ∣}$ $E x c i r c l e s w i t h e x c e n t e r s P, Q, R :$ $r_{P} = \sqrt{\frac{s (a - v) (s - w)}{s - u}}$ $r_{Q} = \sqrt{\frac{s (a - u) (s - w)}{s - v}}$ $r_{R} = \sqrt{\frac{s (a - u) (s - v)}{s - w}}$ $s i m i l a r l y t o i n c i r c l e w e g e t$ $\Rightarrow x_{P} = \frac{(a \sqrt{c^{2} + d^{2}} - c \sqrt{a^{2} + b^{2}}) r_{P}}{∣ a d - b c ∣}$ $\Rightarrow y_{P} = \frac{(b \sqrt{c^{2} + d^{2}} - d \sqrt{a^{2} + b^{2}}) r_{P}}{∣ a d - b c ∣}$ $\Rightarrow x_{Q} = \frac{(a \sqrt{c^{2} + d^{2}} + c \sqrt{a^{2} + b^{2}}) r_{Q}}{∣ a d - b c ∣}$ $\Rightarrow y_{Q} = \frac{(b \sqrt{c^{2} + d^{2}} + d \sqrt{a^{2} + b^{2}}) r_{Q}}{∣ a d - b c ∣}$ $\Rightarrow x_{R} = \frac{(- a \sqrt{c^{2} + d^{2}} + c \sqrt{a^{2} + b^{2}}) r_{R}}{∣ a d - b c ∣}$ $\Rightarrow y_{R} = \frac{(- b \sqrt{c^{2} + d^{2}} + d \sqrt{a^{2} + b^{2}}) r_{R}}{∣ a d - b c ∣}$
	Commented by ajfour last updated on 11/Nov/18

	$E x c e l l e n t S i r .$ $b e a u t i f u l r e s u l t s .$
	Commented by peter frank last updated on 11/Nov/18

	$what app do you use sir ?$
	Commented by mr W last updated on 11/Nov/18

	Commented by mr W last updated on 11/Nov/18

	$t o d r a w t h e g r a p h I u s e d t h e a p p G r a p h e r .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum