Math  /  Geometry

Question```latex \text{3. Use each vertex with its corresponding altitude's slope to graph the three altitudes of the triangle on the grid in Item 1.} \\ \text{4. After drawing the altitudes, Geoff is surprised to see that all three altitudes meet at one point. State the coordinates of the point of concurrency.} \\
\text{Al is not convinced that the altitudes of the triangle are concurrent and wants to use algebra to determine the coordinates of the points of intersection of the altitudes.} \\ \text{5. To use algebra to find the point where the altitudes meet, we need to know the equations of the altitudes.} \\ \text{a. State the coordinates of one point on the altitude from } A \text{ to } \overline{CG}. \\ \text{b. Write the equation of the altitude from } A \text{ to } \overline{CG}. \\ \text{c. Write the equation of the altitude from } C \text{ to } \overline{AG}. \\ \text{d. Write the equation of the altitude from } G \text{ to } \overline{AC}. \\
\text{Extracted text from attached image:} \\ \text{Al, Geoff, and Cal decide that their first priority is to find the treasure. They begin with the first clue from Leon's poem.} \\
\text{At the point on this map where two altitudes cross, In a hole in the ground some treasure was tossed} \\
\text{Since they do not know which two altitudes Leon meant, Geoff decides to place a grid over the map and draw all three altitudes to find the coordinates of any points of intersection.} \\
\text{Slope formula } a = m \frac{y_{1} - y_{1}}{20} A(0,0) \\ \text{1. Determine the slopes of the three sides of } \triangle AGC. \\ \text{e slopes} \\ \text{ocals.} \\ AC \\ \text{il line} \\ \text{ine is} \\ \qquad \\ \begin{array}{l} AC h = \frac{0-0}{36-0} = \frac{AG}{36} = 0 \\ \frac{1}{24-0} \\ \frac{24}{12-0} = \frac{24}{12} = 2 \end{array} \\ \text{2. Determine the slopes of the altitudes of the triangle.} \\ \text{a. What is the slope of the altitude from } A \text{ to } \overline{CG} \text{? Justify your answer.} \\
A \text{ must be perpendicular to the slope of } CG, \text{ so (1)} \\ \text{b. What is the slope of the altitude from } C \text{ to } \overline{AG} \text{? Justify your answer.} \\ \text{undifined} \\ \text{slope } = 0 \\ \text{c. What is the slope of the altitude from } G \text{ to } \overline{AC} \text{? Justify your answer.} \\ \frac{2}{-\frac{1}{2}} ```

Studdy Solution

STEP 1

What is this asking? We need to find where the lines representing the altitudes of triangle AGCAGC intersect, both by graphing and by using algebra with equations. Watch out! Remember, an *altitude* of a triangle is a line segment from a vertex perpendicular to the opposite side (or to the line containing the opposite side).

STEP 2

1. Graph the altitudes.
2. Find the equation of the altitude from *A*.
3. Find the equation of the altitude from *C*.
4. Find the equation of the altitude from *G*.

STEP 3

We already know the slopes of the altitudes from the previous parts of the problem.
The altitude from AA to CG\overline{CG} has a slope of **1**.
The altitude from CC to AG\overline{AG} has a slope of **0**.
The altitude from GG to AC\overline{AC} has an *undefined* slope, meaning it's a vertical line.

STEP 4

Draw the altitudes on the graph, starting from each vertex and using their respective slopes.
You should see them all intersect at a single point!

STEP 5

We know the altitude from AA passes through the point A(0,0)A(0,0) and has a slope of **1**.

STEP 6

Since we have a point (0,0)(0,0) and the slope m=1m=1, we can use the point-slope form of a linear equation: yy1=m(xx1)y - y_1 = m(x - x_1).
Plugging in our values, we get y0=1(x0)y - 0 = 1(x - 0), which simplifies to y=xy = x.

STEP 7

The altitude from CC passes through C(36,0)C(36,0) and has a slope of **0**.

STEP 8

Using the point-slope form again with (36,0)(36,0) and m=0m=0, we get y0=0(x36)y - 0 = 0(x - 36), which simplifies to y=0y = 0.
This makes sense, as a horizontal line has a slope of zero.

STEP 9

The altitude from GG passes through G(12,24)G(12,24) and has an undefined slope, meaning it's a vertical line.

STEP 10

Vertical lines have equations of the form x=cx = c, where cc is a constant.
Since the line passes through G(12,24)G(12,24), the equation is x=12x = 12.

STEP 11

By graphing, you should see the altitudes intersect at the point (12,12)(12,12).
Algebraically, we can find the intersection of y=xy=x and x=12x=12.
Substituting x=12x=12 into y=xy=x gives us y=12y=12.
Since the third line, y=0y=0, isn't involved in finding this intersection (because it intersects at a different point with each of the other lines), we confirm the intersection point is indeed (12,12)(12,12).

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