The equations of the sides AB, BC and CA of ΔABC are y – x = 2, x +

1.	The equations of the sides AB, BC and CA of ΔABC are y – x = 2, x + 2y = 1 and 3x + y + 5 = 0 respectively. The equation of the altitude through B isA. x – 3y + 1 0B. x – 3y + 4 = 0 C. 3x – y + 2 = 0 D. none of these
Answer» The equation of the sides AB, BC and CA of ∆ABC are y − x = 2, x + 2y = 1 and 3x + y + 5 = 0, respectively. Solving the equations of AB and BC, i.e. y − x = 2 and x + 2y = 1, we get: x = − 1, y = 1 So, the coordinates of B are (− 1, 1). The altitude through B is perpendicular to AC. ∴ Slope of AC = -3 Thus, slope of the altitude through B is 13. Equation of the required altitude is given below: y – 1 = 13x + 1 ⇒ x – 3y + 4 = 0

The equations of the sides AB, BC and CA of ΔABC are y – x = 2, x + 2y = 1 and 3x + y + 5 = 0 respectively. The equation of the altitude through B isA. x – 3y + 1 0B. x – 3y + 4 = 0 C. 3x – y + 2 = 0 D. none of these

Answer»

The equation of the sides AB, BC and CA of ∆ABC are y − x = 2, x + 2y = 1 and 3x + y + 5 = 0, respectively.

Solving the equations of AB and BC, i.e. y − x = 2 and x + 2y = 1, we get:

x = − 1, y = 1

So, the coordinates of B are (− 1, 1).

The altitude through B is perpendicular to AC.

∴ Slope of AC = -3

Thus, slope of the altitude through B is 13.

Equation of the required altitude is given below:

y – 1 = 13x + 1

⇒ x – 3y + 4 = 0