Question

find the desired slopes and lengths, then fill in the words that characterize the triangle.answer attempt 1 out of 2slope of \\( \overline { p q } = \frac { - 7 } { 2 } \\)length of \\( \overline { p q } = \sqrt { 53 } \\)slope of \\( \overline { q r } = \frac { 2 } { 3 } \\)length of \\( \overline { q r } = \sqrt { 13 } \\)slope of \\( \overline { r p } = - 1 \\)length of \\( \overline { r p } = 5 \sqrt { 2 } \\)triangle pqr is

Explanation:

Step1: Calculate slopes

The slope formula is \(m=\frac{y_2 - y_1}{x_2 - x_1}\).
For \(\overline{PQ}\): Let \(P(x_1,y_1)\) and \(Q(x_2,y_2)\). Slope \(m_{PQ}=\frac{y_Q - y_P}{x_Q - x_P}\) (values used to get \(-\frac{7}{2}\)).
For \(\overline{QR}\): Slope \(m_{QR}=\frac{y_R - y_Q}{x_R - x_Q}\) (values used to get \(\frac{2}{3}\)).
For \(\overline{RP}\): Slope \(m_{RP}=\frac{y_P - y_R}{x_P - x_R}\) (values used to get \(- 1\)).

Step2: Calculate lengths

The distance formula is \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\).
For \(\overline{PQ}\): \(d_{PQ}=\sqrt{(x_Q - x_P)^2+(y_Q - y_P)^2}=\sqrt{53}\).
For \(\overline{QR}\): \(d_{QR}=\sqrt{(x_R - x_Q)^2+(y_R - y_Q)^2}=\sqrt{13}\).
For \(\overline{RP}\): \(d_{RP}=\sqrt{(x_P - x_R)^2+(y_P - y_R)^2}=5\sqrt{2}\).

Step3: Check for right - angled

If two lines are perpendicular, the product of their slopes is \(-1\). \(m_{PQ}\times m_{QR}=-\frac{7}{2}\times\frac{2}{3}=-\frac{7}{3}
eq - 1\), \(m_{QR}\times m_{RP}=\frac{2}{3}\times(-1)=-\frac{2}{3}
eq - 1\), \(m_{RP}\times m_{PQ}=(-1)\times(-\frac{7}{2})=\frac{7}{2}
eq - 1\). Also, check Pythagoras: \((\sqrt{13})^2+(\sqrt{53})^2 = 13 + 53=66
eq(5\sqrt{2})^2 = 50\), \((\sqrt{13})^2+(5\sqrt{2})^2=13 + 50 = 63
eq(\sqrt{53})^2=53\), \((\sqrt{53})^2+(5\sqrt{2})^2=53 + 50=103
eq(\sqrt{13})^2 = 13\). All side lengths are different (\(\sqrt{13}
eq\sqrt{53}
eq5\sqrt{2}\)).

Answer:

slope of \(\overline{PQ}=-\frac{7}{2}\), length of \(\overline{PQ}=\sqrt{53}\), slope of \(\overline{QR}=\frac{2}{3}\), length of \(\overline{QR}=\sqrt{13}\), slope of \(\overline{RP}=-1\), length of \(\overline{RP}=5\sqrt{2}\), Triangle \(PQR\) is scalene and not right.