If two opposite vertices of a square are (5, 4) and (1, −6), find the coordinates of its remaining two vertices.
The distance $d$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by the formula
$d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$
In a square all the sides are of equal length. The diagonals are also equal to each other. Also in a square the diagonal is equal to times the side of the square.
Here let the two points which are said to be the opposite vertices of a diagonal of a square be A(5,4) and C(1,−6).
Let us find the distance between them which is the length of the diagonal of the square.
$A C=\sqrt{(5-1)^{2}+(4+6)^{2}}$
$=\sqrt{(4)^{2}+(10)^{2}}$
$=\sqrt{16+100}$
$A C=2 \sqrt{29}$
Now we know that in a square,
Side of the square $=\frac{\text { Diagonal of the square }}{\sqrt{2}}$
Substituting the value of the diagonal we found out earlier in this equation we have,
Side of the square $=\frac{2 \sqrt{29}}{\sqrt{2}}$
Side of the square $=\sqrt{58}$
Now, a vertex of a square has to be at equal distances from each of its adjacent vertices.
Let P(x, y) represent another vertex of the same square adjacent to both ‘A’ and ‘C’.
$A P=\sqrt{(5-x)^{2}+(4-y)^{2}}$
$C P=\sqrt{(1-x)^{2}+(-6-y)^{2}}$
But these two are nothing but the sides of the square and need to be equal to each other.
$A P=C P$
$\sqrt{(5-x)^{2}+(4-y)^{2}}=\sqrt{(1-x)^{2}+(-6-y)^{2}}$
Squaring on both sides we have,
$(5-x)^{2}+(4-y)^{2}=(1-x)^{2}+(-6-y)^{2}$
$25+x^{2}-10 x+16+y^{2}-8 y=1+x^{2}-2 x+36+y^{2}+12 y$
$8 x+20 y=4$
$2 x+5 y=1$
From this we have, $x=\frac{1-5 y}{2}$
Substituting this value of ‘x’ and the length of the side in the equation for ‘AP’ we have,
$A P=\sqrt{(5-x)^{2}+(4-y)^{2}}$
$\sqrt{58}=\sqrt{(5-x)^{2}+(4-y)^{2}}$
Squaring on both sides,
$58=(5-x)^{2}+(4-y)^{2}$
$58=\left(5-\left(\frac{1-5 y}{2}\right)\right)^{2}+(4-y)^{2}$
$58=\left(\frac{9+5 y}{2}\right)^{2}+(4-y)^{2}$
$58=\frac{81+25 y^{2}+90 y}{4}+16+y^{2}-8 y$
$232=81+25 y^{2}+90 y+64+4 y^{2}-32 y$
$87=29 y^{2}+58 y$
We have a quadratic equation. Solving for the roots of the equation we have,
$29 y^{2}+58 y-87=0$
$29 y^{2}+87 y-29 y-87=0$
$29 y(y+3)-29(y+3)=0$
$(y+3)(29 y-29)=0$
$(y+3)(y-1)=0$
The roots of this equation are −3 and 1.
Now we can find the respective values of ‘x’ by substituting the two values of ‘y’
When
$x=\frac{1-5(-3)}{2}$
$=\frac{1+15}{2}$
$x=8$
When $y=1$
$x=\frac{1-5(1)}{2}$
$=\frac{1-5}{2}$
$x=-2$
Therefore the other two vertices of the square are $(8,-3)$ and $(-2,1)$.
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