Question

03)Determine o valor de p, nos seguintes casos:
a) A distância de A (p, p – 1) até B (7, 9) é 15;
b) A distância de M (2 – p, 1) até N (p + 7, 3p + 6) é 25.

Answer 1

$\boxed{d_{PQ}^2=(x_Q-x_P)^2+(y_Q-y_P)^2}$

a)

$A(x_A,y_A)=A(p,p-1)\\ B(x_B,y_B)=B(7,9)\\\\ d_{AB}=15\ \therefore\ d^2_{AB}=15^2\ \therefore\\\\ (7-p)^2+(9-(p-1))^2=225\ \therefore\ (7-p)^2+(10-p)^2=225\ \therefore\\\\ 49-14p+p^2+100-20p+p^2=225\ \therefore\\\\ 2p^2-34p+149-225=0\ \therefore\ p^2-17p-38=0\ \therefore$

$p=\dfrac{-(-17)\pm\sqrt{(-17)^2-4(1)(-38)}}{2(1)}=\dfrac{17\pm\sqrt{441}}{2}=$

$=\dfrac{17\pm21}{2}\ \therefore\ \boxed{p=19}\ \ \text{ou}\ \ \boxed{p=-2}$

b)

$M(x_M,y_M)=M(2-p,1)\\ N(x_N,y_N)=N(p+7,3p+6)\\\\ d_{MN}=25\ \therefore\ d_{MN}^2=25^2\ \therefore\\\\ (p+7-(2-p))^2+(3p+6-1)^2=625\ \therefore\\\\ (2p+5)^2+(3p+5)^2=625\ \therefore\\\\ 4p^2+20p+25+9p^2+30p+25=625\ \therefore\\\\ 13p^2+50p+50-625=0\ \therefore\ 13p^2+50p-575=0\ \therefore$

$p=\dfrac{-50\pm\sqrt{50^2-4(13)(-575)}}{2(13)}=\dfrac{-50\pm\sqrt{32400}}{2(13)}=$

$=\dfrac{-50\pm180}{2(13)}=\dfrac{-25\pm90}{13}\ \therefore\ \boxed{p=5}\ \ \text{ou}\ \ \boxed{p=-\dfrac{115}{13}}$