Consider the point A (0, 0), B (4, 2) and C (8, 0)1. Find the mid-poin

1.	Consider the point A (0, 0), B (4, 2) and C (8, 0)1. Find the mid-point of AB.2. Find the equation of the perpendicular bisector of AB.3. Find the equation of the circum circle (Circle passing through the point A, B, and C) of triangle ABC.
Answer» 1. Mid-point of AB is (2, 1). 2. Slope of line through AB = \(\frac{2-0}{4-0} = \frac{1}{2}\) Slope of perpendicular line is – 2 Equation of the perpendicular line to AB is y – 1 = -2(x – 2) ⇒ 2x + y = 5. 3. The meeting point of perpendicular bisector of AB and AC will be the centre of the circum circle. The line perpendicular to AC is x = 4 Solving and x = 4 We get y = 5 – 8 = -3 and x = 4 Hence center is (4, -3) and radius is \(\sqrt{(4)^2 + (-3)^2} = \sqrt{16 + 9} = 5\) Equation of the circle is (x – 4)² + (y + 3)² = 5.

Consider the point A (0, 0), B (4, 2) and C (8, 0)1. Find the mid-point of AB.2. Find the equation of the perpendicular bisector of AB.3. Find the equation of the circum circle (Circle passing through the point A, B, and C) of triangle ABC.

Answer»

1. Mid-point of AB is (2, 1).

2. Slope of line through AB

= \(\frac{2-0}{4-0} = \frac{1}{2}\)

Slope of perpendicular line is – 2

Equation of the perpendicular line to AB is

y – 1 = -2(x – 2) ⇒ 2x + y = 5.

3. The meeting point of perpendicular bisector of AB and AC will be the centre of the circum circle.

The line perpendicular to AC is x = 4

Solving and x = 4

We get y = 5 – 8 = -3 and x = 4

Hence center is (4, -3) and radius is

\(\sqrt{(4)^2 + (-3)^2} = \sqrt{16 + 9} = 5\)

Equation of the circle is

(x – 4)² + (y + 3)² = 5.

Discussion

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