Question

the points ( l(2,2) ), ( m(-3,3) ), and ( n(-2,8) ) form a triangle. find the desired slopes and lengths, then fill in the words that characterize the triangle.
answer
slope of ( overline{lm} = ) slope of ( overline{mn} = ) slope of ( overline{ln} =
length of ( overline{lm} = ) length of ( overline{mn} = ) length of ( overline{ln} =
triangle ( lmn ) is

Explanation:

Step1: Find the slope of \(\overline{LM}\)

The slope formula is \(m=\frac{y_2 - y_1}{x_2 - x_1}\). For \(L(2,2)\) and \(M(-3,3)\), \(m_{LM}=\frac{3 - 2}{-3 - 2}=\frac{1}{-5}=-\frac{1}{5}\)

Step2: Find the slope of \(\overline{MN}\)

For \(M(-3,3)\) and \(N(-2,8)\), \(m_{MN}=\frac{8 - 3}{-2+3}=\frac{5}{1} = 5\)

Step3: Find the slope of \(\overline{LN}\)

For \(L(2,2)\) and \(N(-2,8)\), \(m_{LN}=\frac{8 - 2}{-2 - 2}=\frac{6}{-4}=-\frac{3}{2}\)

Step4: Find the length of \(\overline{LM}\)

The distance formula is \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). For \(L(2,2)\) and \(M(-3,3)\), \(d_{LM}=\sqrt{(-3 - 2)^2+(3 - 2)^2}=\sqrt{25 + 1}=\sqrt{26}\)

Step5: Find the length of \(\overline{MN}\)

For \(M(-3,3)\) and \(N(-2,8)\), \(d_{MN}=\sqrt{(-2 + 3)^2+(8 - 3)^2}=\sqrt{1+25}=\sqrt{26}\)

Step6: Find the length of \(\overline{LN}\)

For \(L(2,2)\) and \(N(-2,8)\), \(d_{LN}=\sqrt{(-2 - 2)^2+(8 - 2)^2}=\sqrt{16 + 36}=\sqrt{52}=2\sqrt{13}\)

Since \(m_{LM}\times m_{MN}=-\frac{1}{5}\times5=- 1\), \(\overline{LM}\perp\overline{MN}\), and \(d_{LM}=d_{MN}=\sqrt{26}\)

Answer:

slope of \(\overline{LM}=-\frac{1}{5}\), slope of \(\overline{MN}=5\), slope of \(\overline{LN}=-\frac{3}{2}\), length of \(\overline{LM}=\sqrt{26}\), length of \(\overline{MN}=\sqrt{26}\), length of \(\overline{LN}=2\sqrt{13}\), Triangle \(LMN\) is a right - isosceles triangle.