Question

If the line 4x+3y+1=0 cuts the co-ordinate axes at P and Q then the equation of
the perpendicular bisector of PQ is

Answer 1

Answer:

Answer====

4x+3y+1=0

x= -4

y= 5

Answer 2

The equation of  the perpendicular bisector is 96y-72x+7=0 .

Given :

Equation of line 4x+3y+1 = 0 .

It cuts co-ordinate axis at P and Q .

So , co-ordinate of P is $(\dfrac{-1}{4},0)$

and co-ordinate of Q is $(0,\dfrac{-1}{3})$

Therefore , mid point is :

$(\dfrac{\dfrac{-1}{4}+0}{2},\dfrac{0+\dfrac{-1}{3}}{2})\\\\(\dfrac{-1}{8},\dfrac{-1}{6})$

Also , slope of bisector is :

$m_b=\dfrac{-1}{\dfrac{-4}{3}}\\\\m_b=\dfrac{3}{4}$

So , eqation of line is :

$(y-\dfrac{-1}{6})=\dfrac{3}{4}(x-\dfrac{-1}{8})\\\\96y-72x+7=0$

Hence , this is the required solution .

Learn More :

Co-ordinate Geometry

https://brainly.in/question/2657222