Question

1. where is the orthocenter of △abc?

a (-4, 4), b (2, 7), c (2, 1) (coordinates on a coordinate grid)

Explanation:

Step1: Determine the type of triangle

Since \(B(2,7)\) and \(C(2,1)\) have the same \(x -\)coordinate, \(BC\) is a vertical line (\(x = 2\)).
The slope of \(AB\): \(m_{AB}=\frac{7 - 4}{2+4}=\frac{3}{6}=\frac{1}{2}\)
The slope of \(AC\): \(m_{AC}=\frac{1 - 4}{2 + 4}=\frac{-3}{6}=-\frac{1}{2}\)

Step2: Find the altitudes

An altitude from \(A\) to \(BC\) is a horizontal line (because \(BC\) is vertical). The equation of the altitude from \(A(-4,4)\) to \(BC\) is \(y = 4\)
An altitude from \(C\) to \(AB\):
The slope of the altitude from \(C\) to \(AB\) is \(- 2\) (negative reciprocal of \(\frac{1}{2}\)). Using the point - slope form \(y - y_1=m(x - x_1)\) with \(C(2,1)\), \(y-1=-2(x - 2)\), \(y-1=-2x + 4\), \(y=-2x+5\)

Step3: Solve the system of equations of altitudes

We have the system of equations \(

$$\begin{cases}y = 4\\y=-2x + 5\end{cases}$$

\)
Substitute \(y = 4\) into \(y=-2x + 5\): \(4=-2x+5\), \(2x=1\), \(x=\frac{1}{2}\)

Another way:
Since \(BC\) is vertical (\(x = 2\)), the altitude from \(A\) to \(BC\) is \(y = 4\).
Since \(AB\) and \(AC\) are not perpendicular, but we can also note that for a right - angled triangle, the orthocenter is at the right - angle vertex. Here, since \(BC\) is vertical and if we consider the intersection of altitudes.
The altitude from \(B\) to \(AC\):
The slope of \(AC\) is \(-\frac{1}{2}\), so the slope of the altitude from \(B\) to \(AC\) is \(2\). Using point - slope form with \(B(2,7)\), \(y - 7=2(x - 2)\), \(y-7=2x-4\), \(y=2x + 3\)
Intersection of \(y = 4\) (altitude from \(A\)) and \(y = 2x+3\): \(4=2x+3\), \(x=\frac{1}{2}\)

Answer:

The orthocenter of \(\triangle ABC\) is \((2,4)\)