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  • The opposite vertices of a square are <img alt="\mathrm{A(3,4)}" src="https://entra

The opposite vertices of a square \mathrm{A B C D} are \mathrm{A(3,4)} and \mathrm{C(1,-1)}. Then the coordinates of other two vertices are

Option: 1

\mathrm{D\left(\frac{1}{2}, \frac{9}{2}\right), B\left(-\frac{1}{2}, \frac{5}{2}\right)}


Option: 2

\mathrm{D\left(-\frac{1}{2}, \frac{9}{2}\right), B\left(\frac{1}{2}, \frac{5}{2}\right)}


Option: 3

\mathrm{B\left(\frac{9}{2}, \frac{1}{2}\right), D\left(-\frac{1}{2}, \frac{5}{2}\right)}


Option: 4

none of these


Answers (1)

best_answer

Middle point \mathrm{M} of diagonal \mathrm{AC} is

\mathrm{M\left(\frac{3+1}{2}, \frac{4-1}{2}\right)=M\left(2, \frac{3}{2}\right)}

If \mathrm{D} is \mathrm{D(h, k)}\quad \cdots(i)



\mathrm{and \: B\left(x_{1}, y_{1}\right), then \: 2=\frac{x_{1}+h}{2}, \frac{3}{2}=\frac{y_{1}+k}{2}}
\mathrm{\Rightarrow \quad x_{1}=4-h, y_{1}=3-k}
\mathrm{Now, B\left(x_{1}, y_{1}\right)\: is \: B(4-h, 3-k)}\quad \cdots(ii)

Suppose slope of \mathrm{AB} is \mathrm{m} and slope of \mathrm{AC} is \mathrm{\frac{4+1}{3-1}=\frac{5}{2}}

Then \mathrm{\tan \left(45^{\circ}\right)=\left|\frac{m-\frac{5}{2}}{1+\frac{5 m}{2}}\right| \Rightarrow(2 m-5)= \pm(2+5 m)}

\mathrm{\Rightarrow \quad m=-\frac{7}{3}, \frac{3}{7} \Rightarrow Equation of \: A B\: is \: y-4=-\frac{7}{3}(x-3)}
\mathrm{or \: 7 x+3 y-33=0\: and\: equation \: of \: B C\: is \: y+1=\frac{3}{7}(x-1)\: or\: 3x-7y-10= 0 }

solving these two equations we get \mathrm{B\left(\frac{9}{2}, \frac{1}{2}\right) }

\mathrm{\Rightarrow \quad \frac{9}{2}=4-h, \frac{1}{2}=3-k \quad \text { by (ii) } }
\mathrm{\Rightarrow \quad h=-\frac{1}{2}, k=\frac{5}{2} \Rightarrow D(h, k)=\left(-\frac{1}{2}, \frac{5}{2}\right)}


 

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