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

x^{2} + y^{2} + 3 x + 4 y - 3 = 0

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

x^{2} + y^{2} - 3 x + 4 y - 3 = 0

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

x^{2} + y^{2} = 25

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

x^{2} + y^{2} + 6 x + 8 y - 3 = 0

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

Open in App

Solution

The correct option is B $x^{2} + y^{2} - 3 x + 4 y - 3 = 0$

$L_{1} : (x - y - 6) + λ (2 x + y + 3) = 0$
Intersection point of these two lines
$x - y - 6 = 0$ and $2 x + y + 3 = 0$ is $(1, - 5)$
$L_{2} : (x + 2 y - 4) + μ (3 x - 2 y - 4) = 0$
Intersection point of these two lines $x + 2 y - 4 = 0$ and $3 x - 2 y - 4 = 0$ is $(2, 1)$ .
Since lines of these two families are perpendicular to each other,
$∴ (1, - 5)$ and $(2, 1)$ will be the diametric end points of a circle.
Hence, $(x - 1) (x - 2) + (y + 5) (y - 1) = 0$
$\Rightarrow x^{2} + y^{2} - 3 x + 4 y - 3 = 0$ is the locus of their point of intersection.

Suggest Corrections

Similar questions

Q. The locus of the image of the point (2, 3) with respect to the line

(x - 2 y + 3) + λ (2 x - 3 y + 4) = 0 (λ ϵ R)

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. The equation of the circle passing through points of intersection of the circle

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

and the line

x + 2 y = 4

and touches the line

x + 2 y = 0

, 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 $x^{2} + y^{2} - 2 x + 1 = 0$

Q. Consider the family of lines

(x - y - 6) + λ (2 x + y + 3) = 0

and

(x + 2 y - 4) + λ (3 x - 2 y - 4) = 0

. If the lines of these

2

families are at right angle to each other then find the locus of their point of intersection.

Join BYJU'S Learning Program

Consider the family of lines (x−y−6)+λ(2x+y+3)=0 and (x+2y−4)+μ(3x−2y−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

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