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Two vertices of an isosceles triangle are $(2, 0)$ and $(2, 5)$. Find the third vertex if the length of the equal sides is 3.
Given:
Two vertices of an isosceles triangle are $(2, 0)$ and $(2, 5)$.
To do:
We have to find the third vertex if the length of the equal sides is 3.
Solution:
Let the two vertices of an isosceles $\triangle ABC$ be $A (2, 0)$ and $B (2, 5)$ and the co-ordinates of the third vertex be $C(x, y)$.
We know that,
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,
\( \mathrm{AC}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \)
\( =\sqrt{(x-2)^{2}+(y-0)^{2}} \)
\( =\sqrt{(x-2)^{2}+y^{2}} \)
\( \Rightarrow \sqrt{(x-2)^{2}+y^{2}}=3 \)
Squaring on both sides, we get,
\( (x-2)^{2}+y^{2}=9 \)
\( x^{2}-4 x+4+y^{2}=9 \)
\( x^{2}+y^{2}-4 x=9-4=5 \)......(i)
\( \mathrm{BC}=\sqrt{(x-2)^{2}+(y-5)^{2}} \)
\( \sqrt{(x-2)^{2}+(y-5)^{2}}=3 \)
Squaring on both sides, we get,
\( (x-2)^{2}+(y-5)^{2}=9 \)
\( x^{2}-4 x+4+y^{2}-10 y+25=9 \)
\( x^{2}+y^{2}-4 x-10 y=9-4-25=-20 \).......(ii)
Subtracting (ii) from (i), we get,
\( 10 y=25 \)
\( y=\frac{25}{10} \)
\( =\frac{5}{2} \)
Substituting the value of \( y \) in (i), we get,
\( x^{2}-4 x+\left(\frac{5}{2}\right)^{2}=5 \)
\( x^{2}-4 x+\frac{25}{4}-5=0 \)
\( 4 x^{2}-16 x+25-20=0 \)
\( 4 x^{2}-16 x+5=0 \)
\( x=\frac{-(-16) \pm \sqrt{(-16)^{2}-4 \times 4 \times 5}}{2 \times 4} \)
\( =\frac{16 \pm \sqrt{16 \times 11}}{8} \)
\( =\frac{16 \pm 4 \sqrt{11}}{8} \)
\( =\frac{4 \pm \sqrt{11}}{2} \)
\( =2 \pm \frac{\sqrt{11}}{2} \)
Therefore, the co-ordinates of the third vertex are \( \left(2+\frac{\sqrt{11}}{2}, \frac{5}{2}\right) \) or \( \left(2-\frac{\sqrt{11}}{2}, \frac{5}{2}\right) \).
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