If $p_{1}, p_{2}$ are the perpendicular distances from the origin to the two straight lines which are perpendicular to each other, then the locus of the point of intersection of the perpendicular lines is

x^{2} + y^{2} = p_{1} + p_{2}

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x^{2} + y^{2} = {p_{1}}^{2} + {p_{2}}^{2}

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\frac{x_{1}^{2}}{{p_{1}}^{2}} + \frac{y_{1}^{2}}{{p_{2}}^{2}} = 1

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x y = p_{1} p_{2}

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Solution

The correct option is D $x^{2} + y^{2} = {p_{1}}^{2} + {p_{2}}^{2}$
From figure,
As, $A B = O C$ ,
$O A B C$ , is a rectangle ,

From pythagorous,

$O B = \sqrt{(O A)^{2} + (O C)^{2}}$

=> $O B = \sqrt{(p_{1})^{2} + (p_{2})^{2}}$ (1).

From distance formula,
$O B = \sqrt{(h - 0)^{2} + (k - 0)^{2}}$

=> $O B = \sqrt{(h)^{2} + (k)^{2}}$ (2).

From (1) & (2),

$h^{2} + k^{2} = p_{1}^{2} + p_{2}^{2}$

=> $x^{2} + y^{2} = p_{1}^{2} + p_{2}^{2}$

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If p1, p2 are the perpendicular distances from the origin to the two straight lines which are perpendicular to each other, then the locus of the point of intersection of the perpendicular lines is

The correct option is D x2+y2=p12+p22From figure,As, AB=OC,OABC, is a rectangle ,From pythagorous,OB=√(OA)2+(OC)2=> OB=√(p1)2+(p2)2____________(1).From distance formula,OB=√(h−0)2+(k−0)2=> OB=√(h)2+(k)2____________(2).From (1) & (2),h2+k2=p21+p22=> x2+y2=p21+p22

If $p_{1}, p_{2}$ are the perpendicular distances from the origin to the two straight lines which are perpendicular to each other, then the locus of the point of intersection of the perpendicular lines is