If -2-6 is the image of point 4-2 with respect to the line L -0- then equation of line is A- 3 x-2 y-10-0- 6 x-4 y-7-0- 2 x -3 y -5-0D- 3 x -2 y -5-0

The correct option is D 3x 2y+5=0 m_PP'=4 6= 23 ⇒ m_L=32 nbsp; Required line L is the perpendicular bisector of PP' nbsp; So, equation of L=0 is y 4=32(x ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Perpendicular Distance of a Point from a Line

If -2,6 is th...

Question

If $(- 2, 6)$ is the image of point $(4, 2)$ with respect to the line $L = 0$ , then equation of line is

6 x - 4 y - 7 = 0

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

2 x + 3 y - 5 = 0

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

3 x - 2 y + 5 = 0

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

3 x - 2 y + 10 = 0

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

Open in App

Solution

The correct option is C $3 x - 2 y + 5 = 0$

$m_{P P^{'}} = \frac{4}{- 6} = \frac{- 2}{3} \Rightarrow m_{L} = \frac{3}{2}$

Required line $L$ is the perpendicular bisector of $P P^{'}$
So, equation of $L = 0$ is
$y - 4 = \frac{3}{2} (x - 1) \Rightarrow 3 x - 2 y + 5 = 0$

Suggest Corrections

Similar questions

Q. The asymptotes of a hyperbola have center at the point

(1, 2)

and are parallel to the lines

2 x + 3 y = 0

and

3 x + 2 y = 0

. If the hyperbola passes through the point

(5, 3)

, then its equation is

Q. Equation of the line passing through (–1, 1) and perpendicular to the line 2x + 3y + 4 = 0, is

Q. A line passing through point A (-5, -4) meet other three lines x + 3y + 2 = 0, 2x + y + 4 = 0 and

x - y - 5 = 0

at B,C and D respectively If

{(\frac{15}{A B})}^{2} + {(\frac{10}{A C})}^{2} = {(\frac{6}{A D})}^{2}

, then the equation of line is

The equation of a straight line passing through the origin and perpendicular to the straight line 2x + 3y - 7 = 0 is ______.

Q. The equation of the tangent to the parabola

y^{2} = 8 x

inclined at

30^{\circ}

to the x axis is

Join BYJU'S Learning Program

Explore more

Perpendicular Distance of a Point from a Line

Standard XII Mathematics

Join BYJU'S Learning Program

If (−2,6) is the image of point (4,2) with respect to the line L=0, then equation of line is

The correct option is C 3x−2y+5=0 mPP′=4−6=−23⇒mL=32 Required line L is the perpendicular bisector of PP′ So, equation of L=0 is y−4=32(x−1)⇒3x−2y+5=0

If $(- 2, 6)$ is the image of point $(4, 2)$ with respect to the line $L = 0$ , then equation of line is

The correct option is C $3 x - 2 y + 5 = 0$

$m_{P P^{'}} = \frac{4}{- 6} = \frac{- 2}{3} \Rightarrow m_{L} = \frac{3}{2}$

Required line $L$ is the perpendicular bisector of $P P^{'}$
So, equation of $L = 0$ is
$y - 4 = \frac{3}{2} (x - 1) \Rightarrow 3 x - 2 y + 5 = 0$