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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 $(9, 4)$.
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