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  • One diagonal of a square is the intercept of the line <img alt="\mathrm{\frac{x}{a}+\frac{y}{b}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cfrac%7Bx%7D%7Ba%7D&plus;%

One diagonal of a square is the intercept of the line \mathrm{\frac{x}{a}+\frac{y}{b}=1} between the axes. Find the coordinates of other two vertices.

Option: 1

\mathrm{\left(\frac{a+b}{2}, \frac{a+b}{2}\right) \text { and }\left(\frac{b-a}{2}, \frac{a-b}{2}\right)}


Option: 2

\mathrm{\left(\frac{a+b}{2}, \frac{a+b}{2}\right) \text { and }\left(\frac{a-b}{2}, \frac{b-a}{2}\right) }
 


Option: 3

\mathrm{\left(\frac{a-b}{2}, \frac{a-b}{2}\right) \text { and }\left(\frac{a+b}{2}, \frac{a+b}{2}\right)}

 


Option: 4

\mathrm{\left(\frac{b-a}{2}, \frac{b-a}{2}\right) \text { and }\left(\frac{a+b}{2}, \frac{a+b}{2}\right) }


Answers (1)

best_answer

The coordinates of centre \mathrm{E} are \mathrm{\left(\frac{a}{2}, \frac{b}{2}\right)}

Slope of \mathrm{A C=-\frac{b}{a}}

Slope of \mathrm{D B=\frac{a}{b} \Rightarrow \tan \theta=\frac{a}{b} \Rightarrow \sin \theta=\frac{a}{\sqrt{a^2+b^2}},}

\mathrm{ \cos \theta=\frac{b}{\sqrt{a^2+b^2}} }

Using distance form for \mathrm{DB}

\mathrm{ \frac{x-\frac{a}{2}}{\frac{b}{\sqrt{a^2+b^2}}}=\frac{y-\frac{b}{2}}{\frac{a}{\sqrt{a^2+b^2}}}= \pm \frac{1}{2} \sqrt{a^2+b^2} }\mathrm{ \Rightarrow \quad x=\frac{a \pm b}{2}, y=\frac{b \pm a}{2} }

The coordinates of other two vertices are thus \mathrm{B\left ( \frac{a+b}{2},\frac{b+a}{2} \right )}, \mathrm{D\left ( \frac{a-b}{2} ,\frac{b-a}{2}\right )}

Hence option 2 is correct.

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chirag

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