Question

Triangle ABC has vertices A (0,10) B (4,10) and C (-2,4) find the orthocenter of triangle ABC

Answer 1

Answer:

Hence the orthocenter is (-2,12)

Step-by-step explanation:

We need to find the Orthocenter of ΔABC with vertices A(0,10) , B(4,10) and C(-2,4).

" Orthocenter of a triangle is a point of intersection, where three altitudes of a triangle connect ".

Step 1 :  Find the perpendicular slopes of any two sides of the triangle.  

Step 2 : Then by using point slope form, calculate the equation for those two altitudes with their respective coordinates.

Step 1 :  Given coordinates are: A(0,10) , B(4,10) and C(-2,4)

Slope of BC = $\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}= \dfrac{4-10}{-2-4}=\dfrac{-6}{-6}=1$

Perpendicular Slope of BC = -1

( since for two perpendicular lines the slope is given as: $m_{1}\times m_{2}=-1$

where  $m_{1},m_{2}$ are the slope of the two lines. )

Slope of AC = $\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{4-10}{-2-0}=\dfrac{-6}{-2}=3$

Perpendicular Slope of AC= $\dfrac{-1}{3}$

Step 2 :  Equation of AD, slope(m) = -1 and point A = (0,10)

$y - y_{1} = m\times (x-x_{1})$

$y - 10 = -1(x - 0)\\y - 10 = -x \\x + y = 10---------------(1)$

Equation of BE, slope(m) =\dfrac{-1}{3} and point B = (4,10)

$y - y_1 = m(x - x_1)$

$y - 10= \dfrac{-1}{3} \times (x - 4)$

$3y-30=-x+4$

$x+3y =34$----------(2)

Solving equations (1) and (2), we get

(x, y) = (-2,12)

Hence, the orthocenter is (-2,12).

Answer 2

(-2, 12) is the orthocenter of triangle ABC