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).
