Question

Using parametric form, find other two vertices of a square , if the vertices of diagonal of a square are $(3,4)$ and $(1,-1)$.

Answer 1

R.E.F. Image.

length of square each side is equal

using distance formula

$\sqrt{(3-x_1{)}^2+(4-y_1{)}^2}=\sqrt{(x_1-1{)}^2+(y_1-(-1){)}^2}$

$(3-x_1{)}^2+(4-y_1{)}^2=(x_1-1{)}^2+(y_1+1{)}^2$

$\therefore 9-6x_1+x_1^2+16-8y_1+y_1^2=x_1^2-2x_1+1+y_1^2+2y_1+1$

$\therefore -6x_1-8y_1+25=-2x_1+2y_1+2$

$\therefore -6x_1+2x_1-8y_1-2y_1+25-2=0$

$\therefore -4x_1-10y_1+23=0$

$\therefore 4x_1+10y_1-23=0$

$\therefore 4x_1+10y_1=23$

Similar

$4x_2+10y_2=23$

$\to$ As length of sides are equal, we has , from

Pythagoras theorem

$\sqrt{(3-1{)}^2+(4-(-1){)}^2}=\sqrt{2\left[\right(x_2-1{)}^2+(y_2+1{)}^2]}$

$\sqrt{4+25}=\sqrt{2\left[\right(x_2-1{)}^2+(y_2+1{)}^2]}$

$29=2(x_2-1{)}^2+(y_2+1{)}^2$

$29=2\left[\right(\frac{23-10y_2}{4}-1{)}^2+(y_2+1{)}^2]$

$\frac{29}{2}=(\frac{19-10y_2}{4}{)}^2+(y_2+1{)}^2$

$\frac{29}{2}=\frac{{19}^2}{16}+\frac{100y_2^2}{16}-\frac{380y_2}{16}+y_2^2+2y_2+1$

$\frac{29}{2}=22.56+625y_2^2-23.75y_2+y_2^2+2y_2+1$

$\therefore 7.25y_2^2-21.75y_2+22.56+1-145=0$

$725y_2^2-21.75y_2+9.06=0$

$y_2^2-3y_2+1.25=0$

$y_2=2.5$ & 0.5 so, $x_2=-0.5,4.5$ Ans