Find the centroid of the triangle whose sides are 3x – y – 5 = 0, x + 2y – 4 = 0 and 5x + 3y + 1 = 0.
Let the given equations represents the sides BC, CA, and AB respectively of ΔABC.
Find the circumcentre of the triangle formed by the points (2, 1), (1, – 2) and (– 2, 3).
Let A (2, 1), B (1, – 2), C (– 2, 3), D, E, F be the mid points of the sides BC, CA and AB and O be the circumcentre.
Mid point of BC | = | D = ((1 – 2)/2, (– 2 + 3)/2) |
= | (– 1/2, 1/2) | |
Mid point of AB | = | F = ((2 + 1)/2, (1 – 2)/2) |
= | (3/2, – 1/2) | |
Slope of BC | = | (– 2 – 3)/(1 + 2) |
= | – 5/3 | |
⇒ Slope of SD | = | 3/5 |
∴ Equation to the perpendicular bisector SD of BC is | ||
y – 1/2 | = | (3/5)(x + 1/2) |
⇒ 6x – 10y + 8 | = | 0 ------ (1) |
Slope of AB | = | 1 + 2)/(2 – 1) = 3. |
∴ Slope of SF | = | – 1/3 |
∴ Equation to the perpendicular bisector SF of AB is | ||
y + 1/2 | = | (– 1/3)(x – 3/2) |
⇒ 2x + 6y | = | 0 |
⇒ x + 3y | = | 0 -----(2) |