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 (3, 4) and ( – 1, 2).
Answer» Given, The line segment joining the points (3,4) and ( – 1,2) To Find: Find the equation of the line Formula used: The equation of line is (y – y₁) = m(x – x₁) Explanation: Here, The right bisector PQ of AB at C and is perpendicular to AB Now, The coordinate of the mid – points = \([\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}]\) The coordinates of point C are = \([\frac{3-1}{2},\frac{4+2}{2}]\) = (1,3) And, The slope of PQ = \(\frac{1}{Slope\, of \ AB}\) The slope of PQ, m = \(\frac{1}{2-4}\) (-1 - 3) = \(\frac{4}{-2}\) So, The slope of PQ, m = – 2 The required equation of PQ is (y – y₁) = m(x – x₁) y – 3 = – 2(x – 1) y – 3 = – 2x + 2 y + 2x = 5 Hence, The equation of line is y + 2x = 5

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

Answer»

Given, The line segment joining the points (3,4) and ( – 1,2)

To Find: Find the equation of the line

Formula used: The equation of line is (y – y₁) = m(x – x₁)

Explanation: Here, The right bisector PQ of AB at C and is perpendicular to AB

Now, The coordinate of the mid – points = \([\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}]\)

The coordinates of point C are = \([\frac{3-1}{2},\frac{4+2}{2}]\) = (1,3)

And, The slope of PQ = \(\frac{1}{Slope\, of \ AB}\)

The slope of PQ, m = \(\frac{1}{2-4}\) (-1 - 3) = \(\frac{4}{-2}\)

So, The slope of PQ, m = – 2

The required equation of PQ is (y – y₁) = m(x – x₁)

y – 3 = – 2(x – 1)

y – 3 = – 2x + 2

y + 2x = 5

Hence, The equation of line is y + 2x = 5