Respuesta :
Solution
[tex]\begin{gathered} 1)\text{ The slope} \\ P(5,4)\colon x_1=5,_{}y_1\text{= 4} \\ Q(-1,14)\text{ : }x_2=-1,\text{ }y_2\text{= 14 } \\ \text{Slope = }\frac{y_2\text{ - }y_1}{\text{ }x_2\text{ - }x_1\text{ }} \\ \text{Slope = }\frac{14\text{ -4}}{-1-5}\text{ = }\frac{10}{-6}\text{ = -}\frac{5}{2}\text{ = -2}\frac{1}{2} \end{gathered}[/tex][tex]\begin{gathered} 2)\text{ Distance betwe}en\text{ PQ = }\sqrt[]{(y_2-y_1)^2+(x_2-x_1)^2} \\ \text{ PQ = }\sqrt[]{(y_2-y_1)^2+(x_2-x_1)^2} \\ \text{ PQ = }\sqrt[]{(14-4)^2+(-1_{}-5_{})^2} \\ \text{ PQ = }\sqrt[]{(14-4)^2+(-1_{}-5_{})^2} \\ \text{ PQ = }\sqrt[]{(10)^2+(-6_{})^2} \\ \text{ PQ = }\sqrt[]{100\text{ + 36}} \\ \text{ PQ = }\sqrt[]{136} \\ \text{ PQ(D) = 11.66} \\ 3)The\text{ Distance represent the DIAMETER of the circle} \\ 4)\text{ Radius = }\frac{D}{2}\text{ =}\frac{11.66}{2}\text{ = 5.83cm} \\ 5)\text{ Area =}\pi\text{ }\times r^2 \\ \text{Area = }\frac{22}{7}\text{ }\times5.83^2 \\ \text{Area = }\frac{747.75}{7}=106.8cm^2 \end{gathered}[/tex][tex]\begin{gathered} 4)(x,y)\text{ =( }\frac{x1+x2}{2},\text{ }\frac{y1+y2}{2}) \\ \text{ = }\frac{5-1}{2},\frac{14+4}{2} \\ \text{ =(2, 9)} \end{gathered}[/tex]
