Question Statement

Find the equations of the altitudes and the point of concurrency of the triangle $ABC$ when $A(4,-2)$, $B(5,5)$ and $C(-1,3)$. What is the name of the point of concurrency?

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Background and Explanation

An altitude of a triangle is a perpendicular line segment from a vertex to the line containing the opposite side. The three altitudes of any triangle intersect at a single point called the orthocenter. To find the equations of altitudes, we first calculate the slopes of the sides, then use the negative reciprocal to get the slopes of the perpendicular altitudes, and finally apply the point-slope formula.

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Solution

Consider the triangle $ABC$ with vertices $A(4,-2)$, $B(5,5)$ and $C(-1,3)$.

Draw the altitudes $\overline{AE}$, $\overline{BF}$ and $\overline{CD}$ of the triangle. Let $m_1,m_2,m_3$ be the slopes of the sides $\overline{AB}$, $\overline{BC}$ and $\overline{CA}$ of the triangle. Therefore,

$\begin{array}{rl}& m_1=\frac{5-(-2)}{5-4}=\frac{7}{1}=7\\ & m_2=\frac{5-3}{5-(-1)}=\frac{2}{6}=\frac{1}{3}\\ & m_3=\frac{3-(-2)}{-1-4}=\frac{3+2}{-5}=-1\end{array}$

Equation of Altitude $\overline{CD}$ (from $C$ to side $AB$)

Since $\overline{CD}$ is perpendicular to $\overline{AB}$:

$\text{Slope of }\overline{CD}=\frac{-1}{\text{Slope of }\overline{AB}}=\frac{-1}{7}$

Thus, the equation of altitude $\overline{CD}$ passing through $C(-1,3)$ with slope $\frac{-1}{7}$ is:

$\begin{array}{rl}y-3& =\frac{-1}{7}(x-(-1))\text{(Slope-point formula)}\\ 7(y-3)& =-1(x+1)\\ 7y-21& =-x-1\\ x+7y-20& =0\end{array}$

Equation of Altitude $\overline{AE}$ (from $A$ to side $BC$)

Since $\overline{AE}$ is perpendicular to $\overline{BC}$:

$\text{Slope of }\overline{AE}=\frac{-1}{\text{Slope of }\overline{BC}}=\frac{-1}{1/3}=-3$

Thus, the equation of altitude $\overline{AE}$ passing through $A(4,-2)$ with slope $-3$ is:

$\begin{array}{rl}y-(-2)& =-3(x-4)\\ y+2& =-3x+12\\ 3x+y-10& =0\end{array}$

Equation of Altitude $\overline{BF}$ (from $B$ to side $CA$)

Since $\overline{BF}$ is perpendicular to $\overline{CA}$:

$\text{Slope of }\overline{BF}=\frac{-1}{\text{Slope of }\overline{CA}}=\frac{-1}{-1}=1$

Thus, the equation of altitude $\overline{BF}$ passing through $B(5,5)$ with slope $1$ is:

$\begin{array}{rl}y-5& =1(x-5)\\ y-5& =x-5\\ x-y& =0\end{array}$

Finding the Point of Concurrency

To find the point of concurrency of these three altitudes, solve any two of the equations (i), (ii) and (iii) simultaneously. Let us solve equations (ii) and (iii):

From equation (iii): $x=y$

Substitute into equation (ii): $\begin{array}{rl}3y+y-10& =0\\ 4y& =10\\ y& =\frac{5}{2}\end{array}$

Substitute $y=\frac{5}{2}$ into equation (iii): $\begin{array}{rl}x-\frac{5}{2}& =0\\ x& =\frac{5}{2}\end{array}$

Thus, $\left(\frac{5}{2},\frac{5}{2}\right)$ is the point of concurrency of these altitudes.

The point of concurrency of the altitudes of a triangle is called the orthocentre of the triangle.

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Key Formulas or Methods Used

• Slope formula: $m=\frac{y_2-y_1}{x_2-x_1}$
• Perpendicular slope condition: $m_{\perp }=-\frac{1}{m}$ (or $m_1\cdot m_2=-1$)
• Point-slope form: $y-y_1=m(x-x_1)$
• Solving simultaneous linear equations: Substitution or elimination method to find intersection points

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Summary of Steps

1. Calculate slopes of all three sides $AB$, $BC$, and $CA$ using the slope formula.
2. Determine slopes of the altitudes by taking the negative reciprocal of the corresponding side's slope.
3. Write equations for each altitude using the point-slope form with the vertex coordinates and the perpendicular slope:
   • Altitude from $C$: $x+7y-20=0$
   • Altitude from $A$: $3x+y-10=0$
   • Altitude from $B$: $x-y=0$
4. Solve any two altitude equations simultaneously to find their intersection point $\left(\frac{5}{2},\frac{5}{2}\right)$.
5. Identify the point as the orthocenter (the point of concurrency of altitudes).