Question

Qual é o ponto da bissetriz dos quadrantes ímpares no 3º quadrante que dista (raiz quadrada de 20) unidades do ponto (1 , -1) ?

Answer 1

⠀

⠀⠀☞ O ponto que procuramos é (-3, -3), ponto este pertencente a reta y = x que é a bissetriz dos quadrantes ímpares que pertence o terceiro quadrante. ✅

⠀

⠀⠀Vamos inicialmente observar os 4 quadrantes do plano cartesiano

⠀

⠀

$\setlength{\unitlength}{0.95cm}\begin{picture}(6,5)\thicklines\put(0,0){\vector(1,0){5}}\put(0,0){\vector(0,1){5}}\put(0,0){\vector(-1,0){5}}\put(0,0){\vector(0,-1){5}}\put(4.8,0.2){x}\put(0.2,4.8){y}\bezier(-3,0)(-2.77,2.77)(0,3)\bezier(3,0)(2.77,2.77)(0,3)\bezier(-3,0)(-2.77,-2.77)(0,-3)\bezier(3,0)(2.77,-2.77)(0,-3)\put(-1.8,1.5){\Huge\bf{Q_2}}\put(0.9,1.5){\Huge\bf{Q_1}}\put(-1.7,-1.8){\Huge\bf{Q_3}}\put(0.9,-1.8){\Huge\bf{Q_4}}\put(-6,3.5){\dashbox{0.1}(3.4,1){ \sf 90^{\circ} < Q_2 < 180^{\circ} }}\put(3,3.5){\dashbox{0.1}(3.4,1){ \sf 0^{\circ} < Q_1 < 90^{\circ} }}\put(3,-4.2){\dashbox{0.1}(3.4,1){ \sf 270^{\circ} < Q_4 < 360^{\circ} }}\put(-6,-4.2){\dashbox{0.1}(3.4,1){ \sf 180^{\circ} < Q_3 < 270^{\circ} }}\bezier(0.7,0)(0.65,0.65)(0,0.7)\put(0.0,0.67){\line(7,28){0.38}}\bezier(-0.7,0)(-0.65,0.65)(0,0.7)\put(-0.29,0.4){\line(-4,-40){0.38}}\bezier(-0.7,0)(-0.65,-0.65)(0,-0.7)\put(0,-0.3){\line(-4,-22){0.38}}\bezier(0.7,0)(0.65,-0.65)(0,-0.7)\put(1.07,0){\line(-4,-31){0.38}}\put(5.4,-0.2){\LARGE\text{ \mathbf {0^{\circ}} }}\put(-0.4,5.5){\LARGE\text{ \mathbf {90^{\circ}} }}\put(-6.3,-0.1){\LARGE\text{ \mathbf {180^{\circ}} }}\put(-0.5,-5.8){\LARGE\text{ \mathbf {270^{\circ}} }}\put(-3.6,-0.45){\bf{-1}}\put(3.2,-0.45){\bf{1}}\put(0.2,-3.5){\bf{-1}}\put(0.2,3.4){\bf{1}}\end{picture}$

⠀

$\sf (Esta~imagem~n\tilde{a}o~\acute{e}~visualiz\acute{a}vel~pelo~App~Brainly)$

⠀

⠀

⠀⠀A reta bissetriz dos quadrantes ímpares (1 e 3) é a reta que passa pela origem (b = 0) e que tem uma inclinação de 45º (a = tg (45º) = 1):

⠀

$\LARGE\gray{\boxed{\sf\blue{~~y = x~~}}}$

⠀

⠀⠀A distância entre dois pontos é encontrada através da equação

⠀

$\large\red{\boxed{\pink{\boxed{\begin{array}{rcl} & & \\ & \orange{\rm d_{a, b} = \sqrt{(x_{b}- x_{a})^{2} + (y_{b}- y_{a})^{2}}} & \\ & & \\ \end{array}}}}}$

⠀

$\Large\blue{\sf (\sqrt{20})^2 = (x - 1)^{2} + (y - (-1))^{2} }$

⠀

$\LARGE\blue{\sf 20 = (x - 1)^{2} + (y + 1)^{2} }$

⠀

⠀⠀Sendo y = -x então temos

⠀

$\LARGE\blue{\sf 20 = (x - 1)^{2}  + (x + 1)^{2} }$

⠀

$\Large\blue{\sf 20 = x^2 - 2x + 1 + x^2 + 2x + 1 }$

⠀

$\LARGE\blue{\sf 20 = 2x^2 + 2 }$

⠀

$\LARGE\blue{\sf 10 = x^2 + 1 }$

⠀

$\LARGE\blue{\sf 9 = x^2}$

⠀

$\LARGE\blue{\sf \sqrt{x^2} = \pm \sqrt{9}}$

⠀

$\LARGE\blue{\sf x = \pm 3}$

⠀⠀

⠀⠀Concluímos, portanto, que existem dois pontos que distam em √20 do ponto (-1, 1). Como o enunciado nos pede o ponto que está no 3º quadrante então sabemos que x < 0, ou seja

⠀

$\LARGE\gray{\boxed{\sf\blue{~~x = -3~~}}}$

⠀

⠀⠀Com o valor de x podemos agora encontrar y, tendo em vista que y = x

⠀

$\LARGE\gray{\boxed{\sf\blue{~~y = -3~~}}}$

⠀

⠀⠀Portanto o ponto que procuramos é

⠀

$\large\green{\boxed{\rm~~~\gray{P}~\pink{=}~\blue{ (-3, -3) }~~~}}$ ✅

⠀

⠀

⠀

⠀

$\bf\large\red{\underline{\quad\quad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}}$

⠀⠀☀️ Leia mais sobre distância entre dois pontos:

⠀

✈ https://brainly.com.br/tarefa/37997846

$\bf\large\red{\underline{\quad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad}}$✍

⠀

⠀

⠀

⠀

$\bf\large\red{\underline{\quad\quad\qquad\qquad\qquad\qquad\qquad\qquad\qquad}}$☁

⠀⠀⠀⠀☕ $\Large\blue{\text{\bf Bons~estudos.}}$

⠀

($\orange{D\acute{u}vidas\ nos\ coment\acute{a}rios}$) ☄

$\bf\large\red{\underline{\qquad \qquad \qquad \qquad \qquad \qquad \quad }}\LaTeX$✍

❄☃ $\sf(\purple{+}~\red{cores}~\blue{com}~\pink{o}~\orange{App}~\green{Brainly})$ ☘☀

⠀

⠀

⠀

⠀

$\gray{"Absque~sudore~et~labore~nullum~opus~perfectum~est."}$