Question

in the diagram above, points d, e, and f are the mid - points of segments bc, ac, and ab. use the information to complete the table below.
segment\tcoordinate points\tslope\tequation of line
ab\ta(1,1)\tb(4,7)\t\frac{3}{6}=\frac{1}{6}\tx = \frac{1}{2}x+5
bc\tb(4,4)\tc(0,9)
ac\ta(1,1)\tc(0,9)
de\td(2,8)\te(0.5,5)
ef\te(0.5,5)\tf(2.5,4)
fd\tf(2.5,4)\td(2,8)
what could be made about the segments within the triangle and the opposite side? justify your answer

Explanation:

Step1: Recall slope - formula

The slope formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $m=\frac{y_2 - y_1}{x_2 - x_1}$.

Step2: Recall point - slope form

The point - slope form of a line is $y - y_1=m(x - x_1)$, which can be rewritten in slope - intercept form $y=mx + b$.

Step3: Find slope of $AB$

Given $A(1,1)$ and $B(4,7)$, $m_{AB}=\frac{7 - 1}{4 - 1}=\frac{6}{3}=2$. Using point - slope form with point $A(1,1)$ and $m = 2$, we have $y-1=2(x - 1)$, which simplifies to $y=2x-1$.

Step4: Find slope of $BC$

Given $B(4,7)$ and $C(9,0)$, $m_{BC}=\frac{0 - 7}{9 - 4}=\frac{-7}{5}$. Using point - slope form with point $B(4,7)$ and $m=-\frac{7}{5}$, we have $y - 7=-\frac{7}{5}(x - 4)$, which simplifies to $y=-\frac{7}{5}x+\frac{35 + 28}{5}=-\frac{7}{5}x+\frac{63}{5}$.

Step5: Find slope of $AC$

Given $A(1,1)$ and $C(9,0)$, $m_{AC}=\frac{0 - 1}{9 - 1}=-\frac{1}{8}$. Using point - slope form with point $A(1,1)$ and $m=-\frac{1}{8}$, we have $y - 1=-\frac{1}{8}(x - 1)$, which simplifies to $y=-\frac{1}{8}x+\frac{8 + 1}{8}=-\frac{1}{8}x+\frac{9}{8}$.

Step6: Mid - point formula

The mid - point formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$.

Step7: Find mid - point $D$ of $BC$

$D(\frac{4+9}{2},\frac{7 + 0}{2})=D(\frac{13}{2},\frac{7}{2})$.

Step8: Find mid - point $E$ of $AC$

$E(\frac{1+9}{2},\frac{1 + 0}{2})=E(5,\frac{1}{2})$.

Step9: Find mid - point $F$ of $AB$

$F(\frac{1+4}{2},\frac{1 + 7}{2})=F(\frac{5}{2},4)$.

Step10: Analyze relationships

The line segment joining the mid - points of two sides of a triangle is parallel to the third side and half its length. For example, $DE$ is parallel to $AB$, $EF$ is parallel to $BC$, and $FD$ is parallel to $AC$. This is known as the mid - point theorem.

Answer:

Segment	Coordinate Points	Slope	Equation of Line
$BC$	$B(4,7),C(9,0)$	$-\frac{7}{5}$	$y=-\frac{7}{5}x+\frac{63}{5}$
$AC$	$A(1,1),C(9,0)$	$-\frac{1}{8}$	$y=-\frac{1}{8}x+\frac{9}{8}$
$DE$	$D(\frac{13}{2},\frac{7}{2}),E(5,\frac{1}{2})$	$2$ (parallel to $AB$)	$y - \frac{1}{2}=2(x - 5)\Rightarrow y=2x-\frac{19}{2}$
$EF$	$E(5,\frac{1}{2}),F(\frac{5}{2},4)$	$-\frac{7}{5}$ (parallel to $BC$)	$y-\frac{1}{2}=-\frac{7}{5}(x - 5)\Rightarrow y=-\frac{7}{5}x+\frac{37}{10}$
$FD$	$F(\frac{5}{2},4),D(\frac{13}{2},\frac{7}{2})$	$-\frac{1}{8}$ (parallel to $AC$)	$y - 4=-\frac{1}{8}(x-\frac{5}{2})\Rightarrow y=-\frac{1}{8}x+\frac{64 + 5}{16}=-\frac{1}{8}x+\frac{69}{16}$