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Exercise 9.3 · Q12

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

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The right bisector (perpendicular bisector) of a segment is the line perpendicular to the segment at its midpoint. For points (3,4)(3,4) and (−1,2)(-1,2), the midpoint is (1,3)(1,3) and the slope of the segment is 12\frac{1}{2}, so the perpendicular slope is −2-2. The equation is 2x+y−5=02x + y - 5 = 0.

The right bisector of a line segment is also called the perpendicular bisector. It is the line that is perpendicular to the segment and passes through its midpoint. Every point on this line is equidistant from the two endpoints — that’s the geometric property that defines it.

To find its equation, we need two things:

  1. The midpoint of the segment (where the bisector passes through).
  2. The slope of the segment (so we can find the slope of a line perpendicular to it).

Step-by-step solution

1. Find the midpoint of the segment.

The midpoint MM of points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is given by:

M=(x1+x22,y1+y22)M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)

Here, (x1,y1)=(3,4)(x_1, y_1) = (3, 4) and (x2,y2)=(−1,2)(x_2, y_2) = (-1, 2).

So:

M=(3+(−1)2,4+22)=(22,62)=(1,3)M = \left( \frac{3 + (-1)}{2}, \frac{4 + 2}{2} \right) = \left( \frac{2}{2}, \frac{6}{2} \right) = (1, 3)

The right bisector passes through (1,3)(1, 3).

2. Find the slope of the given segment.

Slope mm of the line through (3,4)(3,4) and (−1,2)(-1,2) is:

m=y2−y1x2−x1=2−4−1−3=−2−4=12m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 4}{-1 - 3} = \frac{-2}{-4} = \frac{1}{2}

Tip

A common shortcut: if the slope of the segment is 12\frac{1}{2}, the perpendicular slope is the negative reciprocal, which is −2-2. You don’t need to re-derive the perpendicular condition each time.

3. Find the slope of the perpendicular bisector. …

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