Question

Obter o valor de m, sabendo que o ponto M(3m - 1 , m + 5) está na bissetriz dos quadrantes pares.

Answer 1

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$\Huge\green{\boxed{\rm~~~\gray{m}~\pink{=}~\blue{3 }~~~}}$

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$\green{\rm\underline{EXPLICAC_{\!\!\!,}\tilde{A}O\ PASSO{-}A{-}PASSO\ \ \ }}$✍

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☺lá, Roro, como tens passado nestes tempos de quarentena⁉ E os estudos à distância, como vão⁉ Espero que bem❗ Acompanhe a resolução abaixo. ✌

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☔ Inicialmente  vamos observar os 4 quadrantes do plano cartesiano

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

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

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☔ A reta que está na bissetriz dos quadrantes pares (2 e 4) é justamente a reta y = x. Sabendo disto, podemos lançar as coordenadas do nosso ponto na função da reta e encontrar o valor de m

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➡ $\large\blue{\sf m + 5 = 3m - 1 }$

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➡ $\large\blue{\sf m - 3m = -5 - 1 }$

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➡ $\large\blue{\sf -2m = -6 }$

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➡ $\large\blue{\sf m = \dfrac{-6}{-2} }$

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➡ $\large\blue{\sf m = 3 }$

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$\Huge\green{\boxed{\rm~~~\gray{m}~\pink{=}~\blue{3 }~~~}}$ ✅

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

☕ $\bf\Large\blue{Bons\ estudos.}$

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

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

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

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

Answer 2

Resposta:

gxvx

Explicação passo-a-passo: