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Question Number 188998 by SLVR last updated on 10/Mar/23

Commented by SLVR last updated on 10/Mar/23

$k i n d l y h e l p m e$
Commented by manxsol last updated on 12/Mar/23

Commented by manxsol last updated on 12/Mar/23

	Answered by a.lgnaoui last updated on 10/Mar/23

	Commented by SLVR last updated on 15/Mar/23

	$s i r . . . e x p l a n a t i o n i n e n g l i s h . .$ $w h e r e i c a n n o t r e m a i n i n g$
	Answered by manxsol last updated on 12/Mar/23

	$p t o g e n e r i c C^{'} (m e d i a n)$ $(x, y, z) = (2 t, t, 2 t)$ $p t o g e n e r i c B (b i s e c t)$ $(x, y, z) = (s, 2 s, 3 s)$ $\frac{A + B}{2} = C^{'} (A C^{'} B) c o l i n e a l$ $A + B = 2 C^{'}$ $(- 1, 1, 1) + (s, 2 s, 3 s) = (4 t, 2 t, 4 t)$ $- 1 + s = 4 t I s = 4 t + 1$ $1 + 2 s = 2 t I I$ $1 + 3 s = 4 t I I I$ $I I I - I I s = 2 t$ $2 t = 4 t + 1$ $t = \frac{- 1}{2} s = - 1$ $B = (- 1, - 2, - 3)$ $C^{'} = (- 2, - 1, - 2)$ $A = (- 1, 1, 1)$ $a n g u l o s$ $B A = (0, 3, 4)$ $b i s e c = (1, 2, 3)$ $B A . b i s e c =∣ B A ∣∣ b i s e c ∣∣ c o s θ$ $0.1 + 3.2 + 4.3 = 5 \sqrt{14} c o s θ$ $c o s θ = \frac{18}{5 \sqrt{14}}$ $B C = C - B$ $B C = (2 t, t, 2 t) + (1, 2, 3)$ $B C = (2 t + 1, t + 2, 2 t + 3)$ $B C ∙ b i s e c t =∣ B C ∣∣ b i s e c t ∣ \frac{18}{5 \sqrt{14}}$ $(2 t + 1, t + 2, 2 t + 3) . (1, 2, 3) =$ $\sqrt{9 t^{2} + 14 + 20 t} . \sqrt{14} \frac{18}{5 \sqrt{14}}$ $2 t + 1 + 2 t + 4 + 6 t + 9 = \frac{18 \sqrt{9 t^{2} + 14 + 20 t}}{5}$ $(10 t + 14) (5) = 18 \sqrt{9 t^{2} + 20 t + 14}$ $(5 t + 7) (5) = 9 \sqrt{9 t^{2} + 14 + 20 t} .$ $625 t^{2} + 1750 t + 1225 = 729 t^{2} + 1134 + 1620 t$ $104 t^{2} - 130 t - 91 = 0$ $t = - 0.5$ $t = 1.75$ $C = (2 t, t, 2 t)$ $C = (- 1, - \frac{1}{2}, - 1)$ $C = (\frac{7}{2}, \frac{7}{4}, \frac{7}{2})$ $B C = (2 t + 1, t + 2, 2 t + 3)$ $B C = (\frac{9}{2}, \frac{15}{4}, \frac{13}{2})$ $B C = (18, 15, 26)$ $L_{B C} :$ $(x, y, z) = (- 1, - 2, - 3) + t (18, 15, 26)$ $x + 1 = 18 t$ $y + 2 = 15 t$ $z + 3 = 26 t$ $\frac{x + 1}{18} = \frac{y + 2}{15} = \frac{z + 3}{26}$ $f o o t p e r p e r d i c u l a r P (s, 2 s, 3 s)$ $P A ∙ P B = 0$ $P A = (- 1, 1, 1) - (s, 2 s, 3 s)$ $P B = (- 1, - 2, - 3) - (s, 2 s, 3 s)$ $[- 1 - s, 1 - 2 s, 1 - 3 s) ∙ (- 1 - s, - 2 - 2 s, - 3 - 3 s$ $(1 + 2 s + s^{2}) + (4 s^{2} + 2 s - 2) + (9 s^{2} + 6 s - 3) = 0$ $14 s^{2} + 10 s - 4 = 0$ $7 s^{2} + 5 s - 2 = 0$ $7 s - 2$ $1 s + 1$ $s = - 1$ $s = \frac{2}{7}$ $P = (- 1, - 2, - 3) i s P t o B$ $P = (\frac{2}{7}, \frac{4}{7}, \frac{6}{7})$ $g r a p h i c y c o n c l u s i o n s$
	Commented by SLVR last updated on 15/Mar/23

	$s i r i c o u l d n o t f o l l o w w e l l .$ $w h i c h o p t i o n (s) c o r r e c t ?$ $i c a n f o l l o w e n g l i s h o n l y$
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