Q4 of 319 Page 24

Find the equations of the altitudes of a ΔABC whose vertices are A (1, 4), B(-3, 2) and C(-5, -3).

Given: The vertices of ∆ABC are A (1, 4), B (− 3, 2) and C (− 5, − 3).

To find:

The equations of the altitudes of a ΔABC whose vertices are A (1, 4), B (-3, 2) and C (-5, -3).

Explanation:

Diagram:

Slope of AB

Slope of BC

Slope of CA

Thus, we have:

Slope of CF = -2

Slope of AD

Slope of BE

Hence,

Equation of CF is: y + 3 = -2(x + 5)

⇒ 2x + y + 13 = 0

Equation of AD is: y – 4(x – 1)

⇒ 2x + 5y – 22 = 0

Equation of BE is: y – 2(x + 3)

⇒ 6x + 7y + 4 = 0

Find the equation of a line passing through (3, -2) and perpendicular to the line x – 3y + 5 = 0.

Find the equation of the perpendicular bisector of the line joining the points (1, 3) and (3, 1).

Find the equation of a line which is perpendicular to the line and which cuts off an intercept of 4 units with the negative direction of y-axis.

If the image of the point (2, 1) with respect to a line mirror is (5, 2), find the equation of the mirror.