Question #139764

2. The coordinates of the vertices of ΔARC are A(3,3), R(1,-1), and C(-2,1). Angela is trying to determine whether this shape is a right triangle or not. She has made a mistake in the work below the graph. Explain the mistake and show the correct solution.

Hint: The grid is provided for your use (optional). The graph is not worth any points and will not be considered when graded.

Hint: When explaining the mistake, make sure you identify how it is wrong AND how it needs to be corrected.


(12 points total)

 Angela’s solution: 

slope of AR = 

slope of RC = 

slope of AC = 

Since AR and RC are opposite reciprocal slopes, those two sides are perpendicular, and the triangle is a right triangle.




Answer:






1
Expert's answer
2020-10-25T18:54:27-0400

Denote the slope of the line segmentjoining pointsAandRbyARs,AandCbyACsandRandCbyRCsARs=3(1)31=42=2ACs=313(2)=25RCs=1(1)21=23By the condition of perpendiicularity, if the slopeof two lines arem1andm2,they are said to be perpendicular ifm1m2=1The mistake was that Angela mistakened3for2in calculating the slope of lineRCIf2was replaced with3,RCs=24=12RCswould have been12=1ARsand as such,RCwould be perpendicular toARmaking the triangle a right-angled triangle.For triangleARCto be a right-angled triangle,one of the line segments joining the pointshas to be perpendicular to the other.ARsRCs1,ARsACs1,&ACsRCs1Therefore, since none of the slopes are oppositereciprocal of the other, the triangle isnot a right-angled triangle.\textsf{Denote the slope of the line segment}\\ \textsf{joining points}\, A \, \textsf{and}\, R \, \textsf{by}\, AR_s,\\ A \, \textsf{and}\, C \, \textsf{by}\, AC_s \, \textsf{and}\, R \, \textsf{and}\, C \, \textsf{by}\, RC_s\\ \begin{aligned} AR_s &= \frac{3 - (-1)}{3 - 1} = \frac{4}{2} = 2\\ AC_s &= \frac{3 - 1}{3 - (-2)} = \frac{2}{5}\\ RC_s &= \frac{1 - (-1)}{-2 - 1} = \frac{-2}{3} \end{aligned}\\ \textsf{By the condition of perpendiicularity, if the slope}\\ \textsf{of two lines are}\, m_1 \, \textsf{and}\, m_2,\\ \textsf{they are said to be perpendicular if}\, m_1 m_2 = -1\\ \textsf{The mistake was that Angela mistakened} -3 \, \textsf{for}\, -2 \\ \textsf{in calculating the slope of line}\, RC \\ \textsf{If}\, -2 \, \textsf{was replaced with}\, -3, RC_s = \frac{-2}{4} = \frac{-1}{2}\\ RC_s \, \textsf{would have been}\, \frac{-1}{2} = \frac{-1}{AR_s} \\ \textsf{and as such,}\, RC \, \textsf{would be perpendicular to}\,AR \\ \textsf{making the triangle a right-angled triangle.}\\ \textsf{For triangle} \, ARC \, \textsf{to be a right-angled triangle},\\ \textsf{one of the line segments joining the points}\\ \textsf{has to be perpendicular to the other.}\\ AR_s RC_s \neq -1, AR_s AC_s \neq -1,\, \&\, AC_s RC_s \neq -1 \\ \textsf{Therefore, since none of the slopes are opposite}\\ \textsf{reciprocal of the other, the triangle is}\\ \textsf{not a right-angled triangle.}


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