Question

the line 2x+3y+12=0 cuts the axes at A and B , then the equation of perpendicular bisector of AB is​

Answer 1

Step-by-step explanation:

Let the line intersects x-axis at A(x, 0) and y-axis at B(y, 0)

Therefore, x- intercept: (put y=0 in the given equation of line)

2x+3y+12=0

2x +3*0+12=0

2x +12 =0

2x = - 12

x = - 6

Hence, A(x, 0) = A(-6, 0)

Now, y- intercept: (put x=0 in the given equation of line)

2x+3y+12=0

2*0 +3y+12=0

3y +12 =0

3y = - 12

y = - 4

Hence, B(0, y) = B(0, - 4)

Let P (x, y) be any point on the perpendicular bisector of AB.

Therefore, PA = PB

${PA}^{2} = { PB}^{2} \\ ( {x + 6})^{2} + ( {y - 0})^{2} = ( {x - 0})^{2} + ( {y + 4})^{2} \\ ( {x + 6})^{2} + ( {y})^{2} = ( {x })^{2} + ( {y + 4})^{2} \\ {x}^{2} + 12x + 36 + {y}^{2} = {x}^{2} + {y}^{2} + 8y + 16 \\ 12x + 36 = 8y + 16 \\ 12x - 8y + 36 - 16 = 0 \\ 12x - 8y + 20 = 0 \\ 3x - 2y + 5 = 0 \: is \: the \: required \\ \: equation \: of \: perpendicular \: bisector \: of \: AB.$

Answer 2

Answer:

First keep y=0 and then u get x as -6 so the point becomes (-6,0), then next keep x=0 and then u get y as -4 and the point becomes (0,-4).

Know think the point as p(x,y),

So PA=PB

DO S.O.B.S

So (x+6)^2+( y-0)^2=( x -0)^2+( y+4)^2

If you simplify u get 3x-2y+5=0

Step-by-step explanation:

I hope u undersrand and please give me thanks