Question

find the coordinates of the orthocenter of a triangle with vertices at each set of points on a coordinate plane.
a. (0,0), (16,4), (4,6)
b. (5,3), (9,7), (6,10)
a. the orthocenter is (2.2,13.2)
(type an ordered pair. simplify your answer.)
b. the orthocenter is
(type an ordered pair. simplify your answer)

Explanation:

Part b:

Step 1: Find the slope of two sides

Let the points be \( A(5,3) \), \( B(9,7) \), \( C(6,10) \).
Slope of \( AB \): \( m_{AB}=\frac{7 - 3}{9 - 5}=\frac{4}{4} = 1 \).
Slope of \( AC \): \( m_{AC}=\frac{10 - 3}{6 - 5}=\frac{7}{1}=7 \).

Step 2: Find the slopes of the altitudes

The altitude from \( C \) to \( AB \) is perpendicular to \( AB \), so its slope \( m_{\perp AB}=-1 \) (since perpendicular slope is negative reciprocal of \( 1 \)).
The altitude from \( B \) to \( AC \) is perpendicular to \( AC \), so its slope \( m_{\perp AC}=-\frac{1}{7} \) (negative reciprocal of \( 7 \)).

Step 3: Write the equations of the altitudes

Equation of altitude from \( C(6,10) \) with slope \( -1 \):
Using point - slope form \( y - y_1=m(x - x_1) \), we have \( y - 10=-1(x - 6) \), which simplifies to \( y=-x + 16 \).

Equation of altitude from \( B(9,7) \) with slope \( -\frac{1}{7} \):
Using point - slope form \( y - 7=-\frac{1}{7}(x - 9) \).
Multiply through by \( 7 \): \( 7y-49=-x + 9 \), so \( x+7y=58 \).

Step 4: Solve the system of equations

We have the system:
\(

$$\begin{cases}y=-x + 16\\x + 7y=58\end{cases}$$

\)
Substitute \( y=-x + 16 \) into \( x + 7y=58 \):
\( x+7(-x + 16)=58 \)
\( x-7x + 112=58 \)
\( - 6x=58 - 112=-54 \)
\( x = 9 \)

Substitute \( x = 9 \) into \( y=-x + 16 \): \( y=-9 + 16 = 7 \). Wait, that can't be right. Wait, let's recalculate the slope of \( AB \) and \( AC \) again.

Wait, slope of \( AB \): \( A(5,3) \), \( B(9,7) \), \( m_{AB}=\frac{7 - 3}{9 - 5}=\frac{4}{4}=1 \) (correct). Slope of \( BC \): \( B(9,7) \), \( C(6,10) \), \( m_{BC}=\frac{10 - 7}{6 - 9}=\frac{3}{-3}=-1 \). Slope of \( AC \): \( A(5,3) \), \( C(6,10) \), \( m_{AC}=\frac{10 - 3}{6 - 5}=7 \) (correct).

Let's take altitude from \( A \) to \( BC \). Slope of \( BC=-1 \), so slope of altitude from \( A \) is \( 1 \) (negative reciprocal of \( - 1 \)). Equation of altitude from \( A(5,3) \) with slope \( 1 \): \( y - 3=1\times(x - 5) \), so \( y=x - 2 \).

Equation of altitude from \( C \) to \( AB \): slope of \( AB = 1 \), so slope of altitude is \( - 1 \), equation: \( y-10=-1(x - 6)\Rightarrow y=-x + 16 \) (correct).

Now solve \( y=x - 2 \) and \( y=-x + 16 \):
Set \( x - 2=-x + 16 \)
\( 2x=18\Rightarrow x = 9 \), then \( y=9 - 2 = 7 \). Wait, but if we check the altitude from \( B \) to \( AC \):

Slope of \( AC = 7 \), so slope of altitude from \( B \) is \( -\frac{1}{7} \). Equation: \( y - 7=-\frac{1}{7}(x - 9) \). When \( x = 9 \), \( y=7 \), which is point \( B \). That means the triangle is a right triangle? Wait, let's check the slopes: \( m_{AB}=1 \), \( m_{BC}=-1 \), so \( AB\perp BC \) (since \( 1\times(-1)=-1 \)). So the right angle is at \( B \). In a right triangle, the orthocenter is at the right - angled vertex. So the orthocenter is \( (9,7) \).

Answer:

\((9,7)\)