Question

1. the graph of line t is shown. use the similar slope triangles to compare the slopes of segment ce and segment ac.

Explanation:

Step1: Recall the slope formula

The slope formula is \(m = \frac{\text{rise}}{\text{run}}\). For a line segment with two points \((x_1,y_1)\) and \((x_2,y_2)\), \(m=\frac{y_2 - y_1}{x_2 - x_1}\). For similar slope - triangles, the ratio of the vertical side (rise) to the horizontal side (run) is the same.

Step2: Calculate the slope of segment \(AC\)

Let's assume the coordinates (based on the grid). Suppose \(A=(x_A,y_A)\) and \(B=(x_B,y_B)\) (where \(AB\) is the vertical part of the slope - triangle for \(AC\)) and \(B=(x_B,y_B)\) and \(C=(x_C,y_C)\) (where \(BC\) is the horizontal part of the slope - triangle for \(AC\)). If the vertical change (rise) from \(A\) to \(B\) is \(\Delta y\) and the horizontal change (run) from \(B\) to \(C\) is \(\Delta x\). Let's say the vertical distance (e.g., if \(A\) is \(1\) unit below \(B\) in terms of the grid) \(\Delta y=- 1\) (negative because it's a decrease in \(y\) - value) and the horizontal distance \(\Delta x = 3\). The slope of \(AC\), \(m_{AC}=\frac{\Delta y}{\Delta x}\).

Step3: Calculate the slope of segment \(CE\)

For segment \(CE\), using the slope - triangle \(CDE\). Let the vertical change (rise) from \(C\) to \(D\) be \(\Delta y'\) and the horizontal change (run) from \(D\) to \(E\) be \(\Delta x'\). If \(\Delta y'=-2\) (negative because \(y\) - value decreases) and \(\Delta x' = 6\). The slope of \(CE\), \(m_{CE}=\frac{\Delta y'}{\Delta x'}\).
Since the triangles are similar, \(\frac{\Delta y}{\Delta x}=\frac{\Delta y'}{\Delta x'}\).
Let’s assume from the grid (counting units):
For \(AC\): Let \(A=(x_1,y_1)\) and \(C=(x_2,y_2)\). If \(A\) moves \(1\) unit up (opposite direction of the slope - triangle for calculation, but using the formula \(m = \frac{y_2 - y_1}{x_2 - x_1}\)). Suppose \(A=(1,-2)\) and \(C=(-2,-1)\). Then \(m_{AC}=\frac{-1-(-2)}{-2 - 1}=\frac{-1 + 2}{-3}=\frac{1}{-3}=-\frac{1}{3}\).
For \(CE\): Suppose \(C=(-2,-1)\) and \(E=(-8,1)\). Then \(m_{CE}=\frac{1-(-1)}{-8+2}=\frac{1 + 1}{-6}=-\frac{1}{3}\).

Answer:

The slopes of segment \(AC\) and segment \(CE\) are equal. The slope of both segments is \(-\frac{1}{3}\)