Question

find the quadratic equation of the perpendicular bisector of ab, where A and B are the point (3,6) and (-3,4) respectively also fimd its point of intersection with (1) x-axis (2) y-axis

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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)