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  •  If the quadrilateral formed by the lines <img alt="\mathrm{a x+b y+c=0, a^{\prime} x+b^{\prime} y+c=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Ba%20x&plus;b%20ya

 If the quadrilateral formed by the lines \mathrm{a x+b y+c=0, a^{\prime} x+b^{\prime} y+c=0},
\mathrm{ a x+b y+c^{\prime}= 0 }
\mathrm{ a^{\prime} x+b^{\prime} y+c^{\prime}=0} have perpendicular diagonals, then

Option: 1

\mathrm{b^{2}+c^{2}=b^{\prime 2}+c^{\prime 2}}


Option: 2

\mathrm{c^{2}+a^{2}=c^{\prime 2}+a^{\prime 2}}


Option: 3

\mathrm{a^{2}+b^{2}=a^{\prime 2}+b^{\prime 2}}


Option: 4

none of these


Answers (1)

best_answer

Since the diagonals are perpendicular, so the given quadrilateral is a rhombus.

\therefore  Since the diagonals are perpendicular, so the given quadrilateral is a rhombus.

\mathrm{\Rightarrow\left|\frac{c^{\prime}-c}{\sqrt{a^{2}+b^{2}}}\right|=\left|\frac{c^{\prime}-c}{\sqrt{a^{\prime 2}+b^{\prime 2}}}\right| \Rightarrow a^{2}+b^{2}=a^{\prime 2}+b^{\prime 2}}

Posted by

Pankaj Sanodiya

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