Find the equation of the right bisector of the line segment joining th

1.	Find the equation of the right bisector of the line segment joining the points (a, b) and (a1, b1).
Answer» Given: A (a, b) and B (a₁, b₁) be the given points To find: Equation of the right bisector of the line segment joining the points (a, b) and (a₁, b₁). Explanation: Let C be the midpoint of AB. ∴ coordinates of C = \(\Big(\frac{a+a_2}{2},\frac{b+b_1}{2}\Big)\) And, slope of AB = \(\frac{b_1-b}{a_1-a}\) So, the slope of the right bisector of AB is \(-\frac{a_1-a}{b_1-b}\) Thus, the equation of the right bisector of the line segment joining the points (a, b) and (a₁, b₁) is \(y - \frac{b+b_1}{2}\) = \(-\frac{a_1-a}{b_1-b}\Big(x-\frac{a+a_1}{2}\Big)\) ⇒ 2 (a₁ - a)x + 2y(b₁- b) + (a²+ b²) – (a₁² + b₁²) = 0 Hence, equation of the required line 2 (a₁ – a)x + 2y(b₁- b) + (a² + b²) – (a₁² + b₁²) = 0

Find the equation of the right bisector of the line segment joining the points (a, b) and (a1, b1).

Answer»

Given: A (a, b) and B (a₁, b₁) be the given points

To find:

Equation of the right bisector of the line segment joining the points (a, b) and (a₁, b₁).

Explanation:

Let C be the midpoint of AB.

∴ coordinates of C = \(\Big(\frac{a+a_2}{2},\frac{b+b_1}{2}\Big)\)

And, slope of AB = \(\frac{b_1-b}{a_1-a}\)

So, the slope of the right bisector of AB is \(-\frac{a_1-a}{b_1-b}\)

Thus, the equation of the right bisector of the line segment joining the points (a, b) and (a₁, b₁) is

\(y - \frac{b+b_1}{2}\) = \(-\frac{a_1-a}{b_1-b}\Big(x-\frac{a+a_1}{2}\Big)\)

⇒ 2 (a₁ - a)x + 2y(b₁- b) + (a²+ b²) – (a₁² + b₁²) = 0

Hence, equation of the required line 2 (a₁ – a)x + 2y(b₁- b) + (a² + b²) – (a₁² + b₁²) = 0