Question

if two opposite vertices of a square are 5,4) and (-1,6); find the co-ordinates of the remaining vertices.​

Answer 1

Answer:

The co-ordinates points of the remaining vertices are $(-2,1)$ and $(8,-3).$

Step-by-step explanation:

Given: Two opposite vertices of a square are $(5,4)$ and $(-1,6).$

To find: The co-ordinates of the remaining vertices.

Solution: Let $ABCD$ be the given square (see below figure) and $A(5,4)$ and $C(-1,6)$ are the opposite vertices.

Let the co-ordinates of $B$ is $(x, y).$

Now, Join $AC.$

This implies, $AB=BC=CD=DA$

The distance between two points $\mathrm{A}\left(x_{1}, y_{1}\right)$ and $\mathrm{B}\left(x_{2}, y_{2}\right)$ is $\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

Therefore, $\mathrm{AB}=\mathrm{BC}$

Now, Squaring on both the sides,

                                $\mathrm{AB}^{2}=\mathrm{BC}^{2}$

             $(x-5)^{2}+(y-4)^{2}=(x-1)^{2}+(y+6)^{2}$

$x^{2}-10 x+25+y^{2}-8 y+16=x^{2}-2 x+1+y^{2}+12 y+36$

         $-10 x+2 x-8 y-12 y=37-41$

                         $-8 x-20 y=-4$

                      $-4(2 x+5 y)=-4$

                             $2 x+5 y=1$

                                    $2 x=1-5 y$

                                      $x=\frac{1-5 y}{2}$

$ABC$ is a right angled triangle.

                 $\mathrm{AC}^{2}=\mathrm{AB}^{2}+\mathrm{BC}^{2}$

$(5-1)^{2}+(4+6)^{2}=x^{2}-10 x+25+y^{2}-8 y+16+x^{2}-2 x+1+y^{2}+12 y+36$

       $(4)^{2}+(10)^{2}=2 x^{2}+2 y^{2}-12 x+4 y+78$

           $16+100=2 x^{2}+2 y^{2}-12 x+4 y+78$

$2 x^{2}+2 y^{2}-12 x+4 y+78-16-100=0$

              $2 x^{2}+2 y^{2}-12 x+4 y-38=0$

                   $x^{2}+y^{2}-6 x+2 y-19=0$

Substituting $x=\frac{1-5 y}{2},$

     $\left(\frac{1-5 y}{2}\right)^{2}+y^{2}-6\left(\frac{1-5 y}{2}\right)+2 y-19=0$

   $\frac{1+25 y^{2}-10 y}{4}+y^{2}-3(1-5 y)+2 y-19=0$

$1+25 y^{2}-10 y+4 y^{2}-12(1-15 y)+8 y-76=0$

   $1+25 y^{2}-10 y+4 y^{2}-12+60 y+8 y-76=0$

                                       $29 y^{2}+58 y-87=0$

                                             $y^{2}+2 y-3=0$

                                        $y^{2}+3 y-y-3=0$

                                   $y(y+3)-1(y+3)=0$

                                          $(y+3)(y-1)=0$

                                                      $y+3=0$   or $y-1=0$

                                                           $y=-3$ or $y=1$

If $y=1,$ then,

  $x=\frac{1-5 y}{2}$

     $\begin{aligned}&=\frac{1-5 \times 1}{2} \\&=\frac{1-5}{2} \\&=\frac{-4}{2} \\&=-2\end{aligned}$

If $y=-3,$ then,

  $\begin{aligned}&x=\frac{1-5(-3)}{2}\end{aligned}$

     $\begin{aligned}&=\frac{1+15}{2} \\&=\frac{16}{2} \\&=8\end{aligned}$

Hence, the co-ordinates points of the remaining vertices are $(-2,1)$ and $(8,-3).$