Question

Question 9
If Q (0, 1) is equidistant from P (5, - 3) and R (x, 6), find the values of x. Also find the distance QR and PR.

Answer 1

PQ = QR(Given)
And distance between the points is given by
$\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2}$
So,
$\sqrt{(5-0{)}^2+(-3-1{)}^2}=\sqrt{(0-x{)}^2+(1-6{)}^2}$
$\sqrt{(5{)}^2+(-4{)}^2}=\sqrt{(-x{)}^2+(-5{)}^2}$
$\sqrt{25+16}=\sqrt{x^2+25}$
$41=x^2+25$
$16=x^2$
$x=\pm 4$
Therefore, point R is (4, 6) or ( - 4, 6).
When point R is (4, 6),
$PR=\sqrt{(5-4{)}^2+(-3-6{)}^2}=\sqrt{1^2+(-9{)}^2}$
$=\sqrt{1+81}=\sqrt{82}$
$QR=\sqrt{(0-4{)}^2+(1-6{)}^2}=\sqrt{(-4{)}^2+(-5{)}^2}$
$=\sqrt{16+25}=\sqrt{41}$
When point R is (-4,6),
$PR=\sqrt{(5-(-4){)}^2+(-3-6{)}^2}=\sqrt{(9{)}^2+(-9{)}^2}$
$=\sqrt{81+81}=9\sqrt{2}$
$QR=\sqrt{(0-(-4){)}^2+(1-6{)}^2}=\sqrt{(4{)}^2+(-5{)}^2}$
$=\sqrt{16+25}=\sqrt{41}$