Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
An equilateral triangle has two vertices at the points $(3, 4)$, and $(-2, 3)$, find the coordinates of the third vertex.
Given:
An equilateral triangle has two vertices at the points $(3, 4)$, and $(-2, 3)$.
To do:
We have to find the coordinates of the third vertex.
Solution:
Let the two vertices of the equilateral triangle be $A (3,4)$ and $B (-2,3)$ and the third vertex be $C (x, y)$.
This implies,
$AB=BC=CA$
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,
\( AB=\sqrt{(-2-3)^2+(3-4)^2} \)
\( =\sqrt{(-5)^2+(-1)^2} \)
\( =\sqrt{25+1} \)
\( =\sqrt{26} \)
\( BC=\sqrt{(x+2)^2+(y-3)^2} \)
\( C A=\sqrt{(3-x)^{2}+(4-y)^{2}} \)
\( \mathrm{BC}=\mathrm{AB} \)
\( \Rightarrow \sqrt{(x+2)^{2}+(y-3)^{2}}=\sqrt{26} \)
Squaring on both sides, we get,
\( \Rightarrow(x+2)^{2}+(y-3)^{2}=26 \)
\( \Rightarrow x^{2}+4 x+4+y^{2}-6 y+9=26 \)
\( \Rightarrow x^{2}+y^{2}+4 x-6 y+13=26 \)
\( \Rightarrow x^{2}+y^{2}+4 x-6 y=26-13=13 \).........(i)
\( \mathrm{CA}=\mathrm{AB} \)
\( \Rightarrow \sqrt{(3-x)^{2}+(4-y)^{2}}=\sqrt{26} \)
Squaring on both sides, we get,
\( \Rightarrow (3-x)^{2}+(4-y)^{2}=26 \)
\( \Rightarrow 9+x^{2}-6 x+16+y^{2}-8 y=26 \)
\( \Rightarrow x^{2}+y^{2}-6 x-8 y+25=26 \)
\( \Rightarrow x^{2}+y^{2}-6 x-8 y=26-25=1 \)..........(ii)
Subtracting (ii) from (i), we get,
\( 10 x+2 y=12 \)
\( 5 x+y=6 \)
\( y=6-5 x \)
Substituting \( y=6-5x \) in (i), we get,
\( x^{2}+(6-5 x)^{2}+4 x-6(6-5 x)=13 \)
\( \Rightarrow x^{2}+36+25 x^{2}-60 x+4 x-36+30 x-13=0 \)
\( \Rightarrow 26 x^{2}-26 x-13=0 \)
\( \Rightarrow 2 x^{2}-2 x-1=0 \)
\( x=\frac{-(-2) \pm \sqrt{(-2)^{2}-4 \times 2 \times(-1)}}{2 \times 2} \)
\( =\frac{2 \pm \sqrt{4+8}}{4} \)
\( =\frac{2 \pm \sqrt{12}}{4} \)
\( =\frac{2 \pm \sqrt{4 \times 3}}{4} \)
\( =\frac{2 \pm 2 \sqrt{3}}{4} \)
\( =\frac{1 \pm \sqrt{3}}{2} \)
\( x=\frac{1+\sqrt{3}}{2} \) or \( x=\frac{1-\sqrt{3}}{2} \)
If \( x=\frac{1+\sqrt{3}}{2}, \) then,
\( y=6-5 x \)
\( =6-\frac{5(1+\sqrt{3})}{2} \)
\( =\frac{12-5-5 \sqrt{3}}{2} \)
\( =\frac{7-5 \sqrt{3}}{2} \)
If \( x=\frac{1-\sqrt{3}}{2}, \) then,
\( y=6-5 x=6-5 \frac{(1-\sqrt{3})}{2} \)
\( =\frac{12-5+5 \sqrt{3}}{2} \)
\( =\frac{7+5 \sqrt{3}}{2} \)
Hence, the coordinates of the third vertex are \( \left(\frac{1+\sqrt{3}}{2}, \frac{7-5 \sqrt{3}}{2}\right) \) or \( \left(\frac{1-\sqrt{3}}{2}, \frac{7+5 \sqrt{3}}{2}\right) \).
163 Views
