Answer:
Yes, the right angle is angle B
Step-by-step explanation:
we have
![A(-2, 3), B(-3,-6),C(2,-1)](https://tex.z-dn.net/?f=A%28-2%2C%203%29%2C%20B%28-3%2C-6%29%2CC%282%2C-1%29)
Plot the vertices
see the attached figure
we know that
If triangle ABC is a right triangle
then
Applying the Pythagoras Theorem
![AB^{2} =AC^{2}+BC^{2}](https://tex.z-dn.net/?f=AB%5E%7B2%7D%20%3DAC%5E%7B2%7D%2BBC%5E%7B2%7D)
the formula to calculate the distance between two points is equal to
![d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%28y2-y1%29%5E%7B2%7D%2B%28x2-x1%29%5E%7B2%7D%7D)
<em>Find the distance AB</em>
![A(-2, 3), B(-3,-6)](https://tex.z-dn.net/?f=A%28-2%2C%203%29%2C%20B%28-3%2C-6%29)
substitute in the formula
![d=\sqrt{(-6-3)^{2}+(-3+2)^{2}}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%28-6-3%29%5E%7B2%7D%2B%28-3%2B2%29%5E%7B2%7D%7D)
![d=\sqrt{(-9)^{2}+(-1)^{2}}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%28-9%29%5E%7B2%7D%2B%28-1%29%5E%7B2%7D%7D)
![AB=\sqrt{82}\ units](https://tex.z-dn.net/?f=AB%3D%5Csqrt%7B82%7D%5C%20units)
<em>Find the distance BC</em>
![B(-3,-6),C(2,-1)](https://tex.z-dn.net/?f=B%28-3%2C-6%29%2CC%282%2C-1%29)
substitute in the formula
![d=\sqrt{(-1+6)^{2}+(2+3)^{2}}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%28-1%2B6%29%5E%7B2%7D%2B%282%2B3%29%5E%7B2%7D%7D)
![d=\sqrt{(5)^{2}+(5)^{2}}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%285%29%5E%7B2%7D%2B%285%29%5E%7B2%7D%7D)
![BC=\sqrt{50}\ units](https://tex.z-dn.net/?f=BC%3D%5Csqrt%7B50%7D%5C%20units)
<em>Find the distance AC</em>
![A(-2, 3),C(2,-1)](https://tex.z-dn.net/?f=A%28-2%2C%203%29%2CC%282%2C-1%29)
substitute in the formula
![d=\sqrt{(-1-3)^{2}+(2+2)^{2}}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%28-1-3%29%5E%7B2%7D%2B%282%2B2%29%5E%7B2%7D%7D)
![d=\sqrt{(-4)^{2}+(4)^{2}}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%28-4%29%5E%7B2%7D%2B%284%29%5E%7B2%7D%7D)
![AC=\sqrt{32}\ units](https://tex.z-dn.net/?f=AC%3D%5Csqrt%7B32%7D%5C%20units)
Verify the Pythagoras theorem
![(\sqrt{82})^{2} =(\sqrt{32})^{2}+(\sqrt{50})^{2}](https://tex.z-dn.net/?f=%28%5Csqrt%7B82%7D%29%5E%7B2%7D%20%3D%28%5Csqrt%7B32%7D%29%5E%7B2%7D%2B%28%5Csqrt%7B50%7D%29%5E%7B2%7D)
---> is true
therefore
Is a right triangle and the right angle is B
Let the walker’s speed be W, then the biker’s is W+1.5W=2.5W. Difference in speed is 1.5W. Distance=speed times time, so 12.5=1.5W×1.5=2.25W, and W=12.5/2.25=1250/225=50/9=5.56 mph. The biker’s speed is 2.5W=125/9=13.89 mph to 2 Dec places.
(Note: “1.5 times faster than a walker’s speed” could not reasonably mean the biker’s speed=1.5W because this would give W=16.67 mph, which is very fast for a runner, let alone a walker (!), and a biker’s speed of 25mph.)
500(.25)=125 is the answer
Okay? what’s the question here
Well, first, find-out how fast he was going the first time, by dividing 310 by 5. This gives you 62 mph. Now, divide 403 miles by 62 miles per hour in order to find-out how long it would take him to drive 403 miles, if he is constantly doing 62 mph. Your final answer is six-and-a-half hours.