Question

The vertices of a triangle are A(10,4), B(-4,9), and (-2,-1). Find the equation of its altitude which passes through (10,4)

• A. x - 5y + 10 = 0
• B. 5x - y + 20 = 0
• C. 5x - y + 10 = 0
• D. x - 5y + 20 = 0

Answer 1

The correct option is A

x - 5y + 10 = 0

Attitude which passes through A is perpendicular to the line BC.

Slope of the line segment BC=

=$\frac{y_2-y_1}{x_2-x_1}=\frac{-1-9}{-2+4}=\frac{-10}{2}=-5$
Let slope of the attitude be m
Since, attitude and line segment BC is perpendicular
$m_1,m_2=-1$
$(-5),m=-1$
$m=\frac{1}{5}$
Equation of attitude which passes through (10,4) is
$y-y_1=m(x-x_1)$
$y-4=\frac{1}{5}(x-10)$
5y-20=x-10
x-5y+10=0