Question

In a triangleABC, 2x + 3y +1=0, x + 2y - 2 = 0 are the perpendicular bisectors of its sides AB and AC
respectively and if A = (3, 2), then the equation of the side BC is
a)2+y-3=0
b)2-y-3=0
c)2x - y - 2 = 0
d)2x + y - 2 = 0​

Answer 1

Step-by-step explanation:

ANSWER

We have,

The equation of perpendicular bisector of AB is

x−y+5=0

⇒y=x+5......(1)

On comparing that,

y=mx+c

Then,

m=1

The equation of AB passing through the point (1,−2).

Then, equation of line

y−y

1

=m(x−x

1

)

⇒y+2=−1(x−1)

⇒y+2=−x+1

⇒x+y+1=0......(2)

Given equation

x+2y=0......(3)

According to question

From equation (2) and (3) to,

From equation (2)

x+y+1=0

⇒y=−(x+1)

Put in (3) to,

x+2y=0

⇒x−2(x+1)=0

⇒x−2x−2=0

⇒−x−2=0

⇒x=−2

Put in equation (3)

y=1

Then,(x

1

,y

1

)=(−2,1)

From equation (1) and (3) to,

Point of intersection

(x

2

,y

2

)=(

3

−10

,

3

5

)

Then equation of

y−y

1

=

x

2

−x

1

y

2

−y

1

(x−x

1

)

⇒y+2=

3

−10

+2

3

5

−1

(x+2)

⇒y+2=

3

−7

3

2

(x+2)

⇒y+2=

7

−2

(x+2)

⇒7y+14=−2x−4

⇒2x+7y=−4−14

⇒2x+7y=−18