Question

Find y so the distance between (7,y) and (3,3) is 5

Answer 1

$d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$
$5 = \sqrt{(3 - 7)^{2} + (3 - y)^{2}}$
$5 = \sqrt{(-4)^{2} + (-y + 3)^{2}}$
$5 = \sqrt{16 + (-y + 3)(-y + 3)}$
$5 = \sqrt{16 + (-y(-y + 3) + 3(-y + 3)}$
$5 = \sqrt{16 + (-y(-y) - y(3) + 3(-y) + 3(3)}$
$5 = \sqrt{16 + (y^{2} - 3y - 3y + 9)}$
$5 = \sqrt{16 + (y^{2} - 6y + 9)}$
$5 = \sqrt{16 + y^{2} - 6y + 9}$
$5 = \sqrt{y^{2} - 6y + 9 + 16}$
$5 = \sqrt{y^{2} - 6y + 25}$
$25 = y^{2} - 6y + 25$
$0 = y^{2} - 6y$
$0 = y(y) - y(6)$
$0 = y(y - 6)$
$0 = y$    $or$    $0 = y - 6$
$0 = y$    $or$    $6 = y$

Answer 2

You'll need to use pythagoras for this one. we know the distance between the 2 points is 5 (hypotenuse) and that the distance between the x coordinates is 4. to find the distance between the vertical parts of the coordinates we need to do 5^2-4^2 which is 9 and the square root of this is 3. We now know that the vertical distance between the two points is 3, so our y coordinate could be 0 or 6.