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  • Let PQR be a right angled isosceles triangle, right angled at P(2, 1). If the equation of the line QR is 2x + y = 3, then the equation representing the pair of lines&nbsp

Let PQR be a right angled isosceles triangle, right angled at P(2, 1). If the equation of the line QR is 2x + y = 3, then the equation representing the pair of lines PQ and PR is

 

Option: 1

\mathrm{3 x^2-3 y^2+8 x y+20 x+10 y+25=0}


Option: 2

\mathrm{3 \mathrm{x}^2-3 \mathrm{y}^2+8 \mathrm{xy}-20 \mathrm{x}-10 \mathrm{y}+25=0}


Option: 3

\mathrm{3 \mathrm{x}^2-3 \mathrm{y}^2+8 \mathrm{xy}+10 \mathrm{x}+15 \mathrm{y}+20=0}


Option: 4

\mathrm{3 x^2-3 y^2-8 x y-10 x-15 y-20=0}


Answers (1)

best_answer

If m is the slope of side PQ or PR, then

\mathrm{\begin{aligned} & \frac{\mathrm{m}-(-2)}{1+(-2) \mathrm{m}}= \pm \tan \pi / 4, \square \text { slope of } \mathrm{QR} \text { is }-2 \text { and } \angle \mathrm{PQR}=\angle \mathrm{PRQ}=45^{\circ} \\ & \Rightarrow m=3,-1 / 3 \end{aligned}

\mathrm{\therefore }Equations of side PQ and PR are

\mathrm{3 x-y-5=0 \text { and } x+3 y-5=0}

Their combined equation is

\mathrm{\begin{aligned} & (3 x-y-5)(x+3 y-5)=0 \\ & \text { i.e. } 3 x^2-3 y^2+8 x y-20 x-10 y+25=\mathrm{b} \end{aligned}}which is given in (b) 

 

Posted by

avinash.dongre

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