Question

MEE AJUUDEEMM POOR FAAVOOR
(UFMG) Sejam r e s duas retas perpendiculares que se interceptam em P (1, 2). Se Q (-1,6) pertence a uma dessas retas, então a equação da outra reta é:$A) X + 2y - 5 = 0\\B) X - 2y + 3 = 0\\C) 2x - y = 0\\D) 2x + y - 4 = 0\\E) 2x + 2y + 7 = 0$

Answer 1

$\large\boxed{\begin{array}{l}\rm (UFMG)~Sejam~r~e~s~duas~retas~perpendiculares\\\rm que~se~interceptam~em~P(1,2). Se~Q(-1,6)~pertence\\\rm a~uma~dessas~retas,ent\tilde ao~a~equac_{\!\!,}\tilde ao~da~outra~reta~\acute e:\\\tt A)~\sf x+2y-5=0\\\tt B)~\sf x-2y+3=0\\\tt C)~\sf 2x-y=0\\\tt D)~\sf 2x+y-4=0\\\tt E)~\sf 2x+2y+7=0\end{array}}$

$\boxed{\begin{array}{l}\underline{\rm c\acute alculo~do~coeficiente~angular~da~reta~PQ.}\\\sf m=\dfrac{y_Q-y_P}{x_Q-x_P}\\\\\sf m=\dfrac{6-2}{-1-1}\\\sf m=\dfrac{4}{-2}=-2.\\\underline{\rm a~reta~passa~pelo~ponto~P(1,2)}\\\sf y=y_0+m\cdot(x-x_0)\\\sf y=2-2\cdot(x-1)\\\sf y=2-2x+2\\\sf y=-2x+4\longrightarrow reta~PQ.\\\sf a~outra~reta~perpendicular~tem~coeficiente~angular\\\sf m\cdot-2=-1\\\sf 2m=1\\\sf m=\dfrac{1}{2}\end{array}}$

$\boxed{\begin{array}{l}\sf y=mx+b\\\sf 2=\dfrac{1}{2}\cdot1 +b\\\sf b=2-\dfrac{1}{2}\\\sf b=\dfrac{3}{2}\\\sf y=\dfrac{1}{2}x+\dfrac{3}{2}\\\underline{\rm multiplicando~por~2~temos:}\\\sf 2y=x+3\\\sf x-2y+3=0\\\huge\boxed{\boxed{\boxed{\boxed{\sf\maltese~alternativa~B}}}}\end{array}}$