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

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $x^{2} + y^{2} - 4 x - 4 y + 6 = 0$
$y - 1 = m_{1} (x - 3)$ having coordintaes $(3, 1)$ and slope $m_{1}$
$y - 3 = m_{2} (x - 1)$ javing coordinates $(1, 3)$ and slope $m_{2}$
$\Rightarrow m_{1} = \frac{y - 1}{x - 3}$ and $m_{2} = \frac{y - 3}{x - 1}$

Given: The two family of straight lines, cut at right angles to each other.
$\Rightarrow m_{1} m_{2} = - 1$

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

$\Rightarrow (y - 1) (y - 3) = - (x - 3) (x - 1)$
$\Rightarrow y^{2} - 4 y + 3 = - (x^{2} - 4 x + 3)$
$\Rightarrow y^{2} - 4 y + 3 + x^{2} - 4 x + 3 = 0$
$\Rightarrow x^{2} + y^{2} - 4 x - 4 y + 6 = 0$

Thus, the locus of their point of intersection is $x^{2} + y^{2} - 4 x - 4 y + 6 = 0$ .

Suggest Corrections

Similar questions

Q. Consider the family of lines

(x - y - 6) + λ (2 x + y + 3) = 0

and

(x + 2 y - 4) + μ (3 x - 2 y - 4) = 0.

If the lines of these two families are at right angle to each other, then the locus of their point of intersection is

Find the equation of circle which touches $2 x - y + 3 = 0$ and pass through the points of intersection of the line $x + 2 y - 1 = 0$ and the circle $x^{2} + y^{2} - 2 x + 1 = 0$

The equation of the circle which intersects circles $x^{2} + y^{2} + x + 2 y + 3 = 0, x^{2} + y^{2} + 2 x + 4 y + 5 = 0$ and

$x^{2} + y^{2} - 7 x - 8 y - 9 = 0$ at right angle, will be

Q. Find the equation of the circle which passes through the points of intersection of circles

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

and

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

and intersects the circle

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

orthogonally.

Q. The angle between the circles

x^{2} + y^{2} + 4 x + 2 y + 1 = 0

and

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

Join BYJU'S Learning Program

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