Question

find equation of perpendicular bisector of the line segment AB where coordinates of A are ( -3,-1 ) & ( 1,1 )​

Answer 1

Answer:

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Answer 2

Answer:

Equation of perpendicular bisector = 2x + y + 2 = 0

Step-by-step explanation:

Given:

• Line segment AB whose coordinates are (-3, -1) and (1, 1)

To Find:

• Equation of the perpendicular bisector of the line segment AB

Solution:

First finding the equation of the line segment AB.

Given two points on the line, the equation of a line is given by,

$\tt \dfrac{y-y_1}{x-x_1} =\dfrac{y_2-y_1}{x_2-x_1}$

Substitute the data,

$\tt \dfrac{y+1}{x+3} =\dfrac{1+1}{1+3}$

$\tt \dfrac{y+1}{x+3} =\dfrac{1}{2}$

2y + 2 = x + 3

x - 2y + 1 = 0

Therefore equation of line AB is x - 2y + 1 = 0

Let PQ be the perpendicular bisector of the line AB.

Hence point Q will be the midpoint of line AB

Finding coordinates of point Q

Point Q = (-3 + 1)/2, (-1 + 1)/2

⇒ (-1, 0)

Now finding slope of the line AB

AB = x - 2y = -1

-2y = -1 + x

y = -1/-2 + x/2

where -1/-2 is the slope of AB.

Slope of AB = 1/2

Since the line PQ is perpendicular to line AB,

Slope of PQ = -1/slope of AB

Slope of PQ = -1/1/2

= -2

Now the equation of a line when a point on it and slope is given is given by,

$\tt y-y_0=m(x-x_0)$

Substitute the data,

y - 0 = -2( x + 1)

y = -2x - 2

y + 2x + 2 = 0

Therefore equation of the perpendicular bisector is 2x + y + 2 = 0.