Question: Find the desired slopes and lengths, then fill in the words that characterize the triangle. Answer…

Find the desired slopes and lengths, then fill in the words that characterize the triangle. Answer Attempt 1 out of 2 slope of PQ = \frac{1}{3} slope of QR = \frac{-10}{2} slope of RP = \frac{9}{-7} length of PQ = \sqrt{26} length of QR = \boxed{} length of RP = \boxed{} Triangle PQR is \boxed{}

Solution

Step 1 Let us assume the coordinates of the points. Let \( P \) be at the origin \((0, 0)\). Step 2 Given the slope of \( PQ \) is \( \frac{1}{5} \), and the length of \( PQ \) is \( \sqrt{26} \). Thus, if \( Q \) is at \((x, y)\), then: \[ \frac{y}{x} = \frac{1}{5} \& \] \[ \sqrt{x^2 + y^2} = \sqrt{26} \& \] Substituting \( y = \frac{1}{5}x \) into the distance equation: \[ \sqrt{x^2 + \left(\frac{1}{5}x\right)^2} = \sqrt{26} \& \] \[ \sqrt{x^2 + \frac{x^2}{25}} = \sqrt{26} \& \] \[ \sqrt{\frac{26x^2}{25}} = \sqrt{26} \& \] \[ \frac{\sqrt{26}x}{5} = \sqrt{26} \& \] \[ x = 5 \& \] Thus, \( y = \frac{1}{5} \times 5 = 1 \), so \( Q \) is at \((5, 1)\). Step 3 Given the slope of \( QR \) is \( -\frac{10}{3} \). Let \( R \) be at \((a, b)\). Thus, the slope of \( QR \) is: \[ \frac{b - 1}{a - 5} = -\frac{10}{3} \& \] \[ b - 1 = -\frac{10}{3}(a - 5) \& \] \[ b = -\frac{10}{3}a + \frac{50}{3} + 1 \& \] \[ b = -\frac{10}{3}a + \frac{53}{3} \& \] Step 4 Given the slope of \( RP \) is \( -\frac{3}{10} \). Since \( P \) is at \((0, 0)\), the slope of \( RP \) is: \[ \frac{b}{a} = -\frac{3}{10} \& \] \[ b = -\frac{3}{10}a \& \] Step 5 Set the two expressions for \( b \) equal to find \( a \): \[ -\frac{3}{10}a = -\frac{10}{3}a + \frac{53}{3} \& \] Multiply both sides by 30 to eliminate denominators: \[ -9a = -100a + 530 \& \] \[ 91a = 530 \& \] \[ a = \frac{530}{91} \& \] Step 6 Substitute the value of \( a \) back into \( b = -\frac{3}{10}a \): \[ b = -\frac{3}{10} \times \frac{530}{91} \& \] \[ b = -\frac{1590}{910} \& \] \[ b = -\frac{159}{91} \& \] Thus, \( R \) is at \( \left( \frac{530}{91}, -\frac{159}{91} \right) \). Step 7 Find the length of \( QR \): \[ QR = \sqrt{ \left(5 - \frac{530}{91}\right)^2 + \left(1 - \left(-\frac{159}{91}\right) \right)^2 } \& \] \[ QR = \sqrt{ \left( \frac{455}{91} - \frac{530}{91} \right)^2 + \left( \frac{91}{91} + \frac{159}{91} \right)^2 } \& \] \[ QR = \sqrt{ \left( -\frac{75}{91} \right)^2 + \left( \frac{250}{91} \right)^2 } \& \] \[ QR = \sqrt{ \frac{5625}{8281} + \frac{62500}{8281} } \& \] \[ QR = \sqrt{ \frac{68125}{8281} } \& \] \[ QR = \frac{\sqrt{68125}}{91} \& \] Simplifying, \( \sqrt{68125} = 261.25 \), so: \[ QR \approx \frac{261.25}{91} \approx 2.87 \& \] Step 8 Find the length of \( RP \): \[ RP = \sqrt{ \left( \frac{530}{91} - 0 \right)^2 + \left( -\frac{159}{91} - 0 \right)^2 } \& \] \[ RP = \sqrt{ \frac{280900}{8281} + \frac{25281}{8281} } \& \] \[ RP = \sqrt{ \frac{306181}{8281} } \& \] \[ RP = \frac{\sqrt{306181}}{91} \& \] Simplifying, \( \sqrt{306181} \approx 553.31 \), so: \[ RP \approx \frac{553.31}{91} \approx 6.08 \& \] Step 9 Characterize Triangle \( PQR \). All sides \( PQ \), \( QR \), and \( RP \) have different lengths: \[ PQ = \sqrt{26} \approx 5.10 \& \] \[ QR \approx 2.87 \& \] \[ RP \approx 6.08 \& \] Thus, Triangle \( PQR \) is a scalene triangle.