Question

solve this chapter name- distance formula question13

Answer 1

1) P(0.5) ;  Q(5, 10),  R(6, 3)

$If\ A=(x_1,y_1); B=(x_2,y_2);\\ Distance\ AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\\\PQ=\sqrt{(5-0)^2+(10-5)^2}=5\sqrt{2}\\\\QR=\sqrt{(6-5)^2+(3-10)^2}=5\sqrt{2}\\\\RP=\sqrt{(6-0)^2+(3-5)^2}=2\sqrt{10}$

PQ=QR  => ISOSCELES.

2)   P(0,-4)     Q(6,2)     R(3,5)    S(-3, -1)
$PQ=\sqrt{6^2+6^2}=6\sqrt2,\ \ QR=\sqrt{3^2+3^2}=3\sqrt2\\\\RS=\sqrt{6^2+6^2}=6\sqrt2,\ SP=\sqrt{(-3)^2+3^2}=3\sqrt2\\\\PR=\sqrt{3^2+9^2}=9\sqrt{10},\ QS=\sqrt{9^2+3^2}=PR$
PQ = RS and SP=QR  and diagonals are equal.
so rectangle.

3)  A(1, -3)   B(-3, 0)   C(4, 1)
$AB=\sqrt{4^2+(-3)^2}=5,\ BC=\sqrt{7^2+1^2}=5\sqrt2\\\\CA=\sqrt{3^2+4^2}=5,\ CA=AB.\\\\AB^2+CA^2=BC^2$

Hence ABC is a right angle (at A) and isosceles triangle.