$y - 1 = m_{1} (x - 3)$ and $y - 3 = m_{2} (x - 1)$ are two family of straight lines, at right angled to each other. The locus of their point of intersection is?

x^{2} + y^{2} - 2 x - 6 y + 10 = 0

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x^{2} + y^{2} - 4 x - 4 y + 6 = 0

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x^{2} + y^{2} - 2 x - 6 y + 6 = 0

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x^{2} + y^{2} - 4 x - 4 y - 6 = 0

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Solution

The correct option is A $x^{2} + y^{2} - 4 x - 4 y + 6 = 0$
Given $y - 1 = m_{1} (x - 3)$ and $y - 3 = m_{2} (x - 1)$

$m_{1} = \frac{y - 1}{x - 3}, m_{2} = \frac{y - 3}{x - 1}$

As the lines are at right angled to each other.

Slopes product $= - 1$

Hence,

$m_{1} \times m_{2} = - 1$

$\Rightarrow \frac{y - 1}{x - 3} \times \frac{y - 3}{x - 1} = - 1$

$\Rightarrow y^{2} - 4 y + 3 = - (x^{2} - 4 x + 3)$

$\Rightarrow x^{2} + y^{2} - 4 x - 4 y + 6 = 0$

$∴$ the locus of intersection $x^{2} + y^{2} - 4 x - 4 y + 6 = 0$

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y−1=m1(x−3) and y−3=m2(x−1) are two family of straight lines, at right angled to each other. The locus of their point of intersection is?

$y - 1 = m_{1} (x - 3)$ and $y - 3 = m_{2} (x - 1)$ are two family of straight lines, at right angled to each other. The locus of their point of intersection is?