How do I solve for y2 in the distance formula? LOOK: Solve for y2. d = sqrt{(x2-x1)^2 + (y2-y1)^2}
G Guest Guest Sep 23, 2006 #1 How do I solve for y2 in the distance formula? LOOK: Solve for y2. d = sqrt{(x2-x1)^2 + (y2-y1)^2}
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,216 Sep 23, 2006 #2 Square both sides to eliminate radical: \(\displaystyle \L\\d^{2}=(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}\) \(\displaystyle \L\\d^{2}-(x_{2}-x_{1})^{2}=(y^{2}-y_{1})^{2}\) \(\displaystyle \L\\\sqrt{d^{2}-(x_{2}-x_{1})^{2}}=(y_{2}-y_{1})\) \(\displaystyle \L\\\sqrt{d^{2}-(x_{2}-x_{1})^{2}}+y_{1}=y_{2}\)
Square both sides to eliminate radical: \(\displaystyle \L\\d^{2}=(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}\) \(\displaystyle \L\\d^{2}-(x_{2}-x_{1})^{2}=(y^{2}-y_{1})^{2}\) \(\displaystyle \L\\\sqrt{d^{2}-(x_{2}-x_{1})^{2}}=(y_{2}-y_{1})\) \(\displaystyle \L\\\sqrt{d^{2}-(x_{2}-x_{1})^{2}}+y_{1}=y_{2}\)
