Question

a). P(3.2) and QX7.6) are two points. Find 1)the equation of the perpendicular bisector of PQ n) the value of bif(3.b) les on bisector.​

Answer 1

Answer:

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

PA=PB

⇒

(x−3)

2

+(y−6)

2

​

=

(x+3)

2

+(y−4)

2

​

⇒(x−3)

2

+(y−6)

2

=(x+3)

2

+(y−4)

2

⇒x

2

−6x+9+y

2

−12y+36=x

2

+6x+9+y

2

−8y+16

⇒12x+4y−20=0

⇒3x+y−5=0

Hence, the equation of the perpendicular bisector of AB is 3x+y-5=0

(i) We know that the coordinates of any point on x-axis are of the form(x,o). In other words, y-coordinate of every point on x-axis is zero. So, putting y=o in (i),we get

3x−5=0⇒x=

3

5

​

Thus, the perpendicular bisector of AB cuts x-axis at(

3

5

​

,0)

(ii)  The coordinates of any point on y-axis are of the form (o,y). Putting x=0 in (i), we get

       y−5=0⇒y=5

Thus, the perpendicular bisector of AB intersects y-axis at (0,5)

( copy paste from topper )

Step-by-step explanation: