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} = mtimes (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).