Question

similar triangles ( abc ) and ( def ) are shown on the coordinate plane. line ( t ) passes through points ( b ), ( e ), ( f ), and ( c ).
use the similar triangles to explain the relationship between the slopes of ( overline{bc} ) and ( overline{ef} ).
drag a response into each box to explain.
the slope of ( overline{bc} ) equals and the slope of ( overline{ef} ) equals.
since ( \triangle abc ) is similar to ( \triangle def ) and ( overline{bc} ) and ( overline{ef} ) both lie on line ( t ), the slope of ( overline{bc} ) is the slope of ( overline{ef} ).

Explanation:

Step1: Recall the slope formula

The slope of a line segment between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(m = \frac{y_2 - y_1}{x_2 - x_1}\). For \(\overline{BC}\), if we consider the vertical change \(AB\) (in the \(y -\) direction) and horizontal change \(AC\) (in the \(x -\) direction), the slope \(m_{BC}=\frac{AB}{AC}\) (using the right - triangle \(ABC\) as a reference for vertical and horizontal changes). For \(\overline{EF}\), if we consider the vertical change \(DE\) (in the \(y -\) direction) and horizontal change \(DF\) (in the \(x -\) direction), the slope \(m_{EF}=\frac{DE}{DF}\).

Step2: Use the property of similar triangles

Since \(\triangle ABC\sim\triangle DEF\), the ratios of their corresponding sides are equal. That is, \(\frac{AB}{DE}=\frac{AC}{DF}\), which can be rewritten as \(\frac{AB}{AC}=\frac{DE}{DF}\).

Step3: Conclude the relationship between slopes

Because \(m_{BC}=\frac{AB}{AC}\) and \(m_{EF}=\frac{DE}{DF}\), and \(\frac{AB}{AC}=\frac{DE}{DF}\) (from the similarity of \(\triangle ABC\) and \(\triangle DEF\)), we have \(m_{BC}=m_{EF}\). Also, since \(\overline{BC}\) and \(\overline{EF}\) lie on the same line \(t\), their slopes must be equal.

Answer:

The slope of \(\overline{BC}\) equals \(\frac{AB}{AC}\) and the slope of \(\overline{EF}\) equals \(\frac{DE}{DF}\).
The slope of \(\overline{BC}\) is equal to the slope of \(\overline{EF}\).