Question

Determine o valor de m nos seguintes casos:
L (m,m+8), p (-14,8) e dlp =26??​

Answer 1

Resposta:

m = -24 ou m = 10

Explicação passo a passo:

A distância entre os pontos L(m,m+8) e P(-14,8) vale 26

dLP²=(xP-xL)²+(yP-yL)²

L(m,m+8) => xL = m e yL = m+8

P(-14,8)    => xP = -14 e yP = 8

Aplicando a fórmula:

26²=(-14-m)²+[8-(m+8)]²

676=m²+2.14.m+14²+[8-m-8]²

676=m²+28m+196+m²

2m²+28m-480=0 ÷(2)

m²+14m-240=0

$\displaystyle Aplicando~a~f\acute{o}rmula~de~Bhaskara~para~m^{2}+14m-240=0~~e~comparando~com~(a)x^{2}+(b)x+(c)=0,~determinamos~os~coeficientes:~\\a=1{;}~b=14~e~c=-240\\\\C\acute{a}lculo~do~discriminante~(\Delta):&\\&~\Delta=(b)^{2}-4(a)(c)=(14)^{2}-4(1)(-240)=144-(-960)=1156$

Fatorar 1156

1156 | 2

578 | 2

289 | 17

17 | 17

1 | 1

1156 = 2 . 2 . 17 . 17 = 2².17²

√2².17²=√2².√17²=2.17=34

$\displaystyle \\\\C\acute{a}lculo~das~raizes:&\\m^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(14)-\sqrt{1156}}{2(1)}=\frac{-(14)-34}{2}=-24\\\\m^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(14)+\sqrt{1156}}{2(1)}=\frac{-14+34}{2}=10\\\\S=\{-24,~10\}$