Find the equation of the right bisector of the line segment joining the points (3, 4) and (–1, 2).

Asked by Abhisek | 1 year ago |  112

##### Solution :-

The right bisector of a line segment bisects the line segment at 90°.

End-points of the line segment AB are given as A (3, 4) and B (–1, 2).

Let mid-point of AB be (x, y)

x = $$\dfrac{3-1}{2}$$$$\dfrac{2}{2}$$ = 1

y = $$\dfrac{4+2}{2}$$$$\dfrac{6}{2}$$ = 3

(x, y) = (1, 3)

Let the slope of line AB be m1

m1 = $$\dfrac{2-4}{-1-3}$$

$$\dfrac{-2}{-4}$$

$$\dfrac{1}{2}$$

And let the slope of the line perpendicular to AB be m2

m2 =$$\dfrac{-1}{\dfrac{1}{2}}$$

= -2

The equation of the line passing through (1, 3) and having a slope of –2 is

(y – 3) = -2 (x – 1)

y – 3 = – 2x + 2

2x + y = 5

The required equation of the line is 2x + y = 5

Answered by Pragya Singh | 1 year ago

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