# The three vertices of a parallelogram are $(3, 4), (3, 8)$ and $(9, 8)$. Find the fourth vertex.

Given:

The three vertices of a parallelogram are $(3, 4), (3, 8)$ and $(9, 8)$.

To do:

We have to find the fourth vertex.

Solution:

Let ABCD be the parallelogram whose vertices $A, B$ and $C$ are $(3, 4), (3, 8), (9, 8)$ respectively.
Let the coordinates of $D$ be $(x, y)$.
$\mathrm{AB}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$
$=\sqrt{(3-3)^{2}+(8-4)^{2}}$

$=\sqrt{0+(4)^{2}}$
$=\sqrt{16}$

$=4$
$\mathrm{BC}=\sqrt{(3-9)^{2}+(8-8)^{2}}$
$=\sqrt{(-6)^{2}+0}$

$=\sqrt{36}$

$=6$
$\mathrm{CD}=\sqrt{(x-9)^{2}+(y-8)^{2}}$
$\mathrm{DA}=\sqrt{(3-x)^{2}+(4-y)^{2}}$
Opposite sides are equal in a parallelogram.
$\therefore A B=C D$ and $B C=A D$

$C D=\sqrt{(x-9)^{2}+(y-8)^{2}}=4$

Squaring on both sides, we get,

$(x-9)^{2}+(y-8)^{2}=(4)^{2}$

$x^{2}-18 x+81+y^{2}-16 y+64=16$

$x^{2}+y^{2}-18 x-16 y=16-81-64$

$x^{2}+y^{2}-18 x-16 y=-129$......(i)

$\mathrm{AD}=\sqrt{(3-x)^{2}+(4-y)^{2}}=6$

$(3-x)^{2}+(4-y)^{2}=36$

Squaring on both sides, we get,
$9+x^{2}-6 x+16+y^{2}-8 y=36$
$x^{2}+y^{2}-6 x-8 y=36-9-16$
$x^{2}+y^{2}-6 x-8 y=11$......(ii)
Subtracting (i) from (ii), we get,

$12 x+8 y=140$
$3 x+2 y=35$

$2 y=35-3 x$
$y=\frac{35-3 x}{2}$
Substituting the value of $y$ in (ii), we get,
$x^{2}+\left(\frac{35-3 x}{2}\right)^{2}-6 x-8\left(\frac{35-3 x}{2}\right)=11$
$x+\frac{1225+9 x^{2}-210 x}{4}-6 x-140+12 x=11$
$4 x^{2}+1225+9 x^{2}-210 x-24 x-560+48 x=44$
$13 x^{2}-186+621=0$
$13 x^{2}-117 x-69 x+621=0$
$13 x(x-9)-69(x-9)=0$
$(x-9)(13 x-69)=0$
$\Rightarrow x-9=0$ or $13x-69=0$

$x=9$ or $x=\frac{69}{13}$ which is not possible

$\therefore x =9$

This implies,

$y =\frac{35-3 x}{2}$
$=\frac{35-3 \times 9}{2}$

$=\frac{35-27}{2}$

$=\frac{8}{2}$

$=4$

Therefore, the fourth vertex is $(9, 4)$.

Updated on: 10-Oct-2022

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