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δlmn has vertices at l(1,8), m(6,9), and n(9, -5). is δlmn a right tria…

Question

δlmn has vertices at l(1,8), m(6,9), and n(9, -5). is δlmn a right triangle? justify your answer. yes, (overline{lm}) is perpendicular to (overline{mn}). yes, (overline{lm}) is perpendicular to (overline{ln}). yes, (overline{ln}) is perpendicular to (overline{mn}). no, none of the sides are perpendicular.

Explanation:

Step1: Calculate the slopes of \(LM\), \(MN\), and \(LN\)

The slope formula is \(m=\frac{y_2 - y_1}{x_2 - x_1}\).
For \(LM\) with \(L(1,8)\) and \(M(6,9)\):
\(m_{LM}=\frac{9 - 8}{6 - 1}=\frac{1}{5}\)
For \(MN\) with \(M(6,9)\) and \(N(9,- 5)\):
\(m_{MN}=\frac{-5 - 9}{9 - 6}=\frac{-14}{3}\)
For \(LN\) with \(L(1,8)\) and \(N(9,-5)\):
\(m_{LN}=\frac{-5 - 8}{9 - 1}=\frac{-13}{8}\)

Step2: Check the product of slopes for perpendicularity

Two lines are perpendicular if \(m_1\times m_2=-1\).
\(m_{LM}\times m_{MN}=\frac{1}{5}\times\frac{-14}{3}=-\frac{14}{15}
eq - 1\)
\(m_{LM}\times m_{LN}=\frac{1}{5}\times\frac{-13}{8}=-\frac{13}{40}
eq - 1\)
\(m_{MN}\times m_{LN}=\frac{-14}{3}\times\frac{-13}{8}=\frac{91}{12}
eq - 1\)

Answer:

No, none of the sides are perpendicular.