Verified Answer :

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.

Panoyin

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

Related Questions