The formula for point-slope form is y-a=m(x-c), where y and x are variables, a is simply one point in y, m is the slope, and c is simply one point in x. The first thing we can find is m, or the slope. This is found by dividing the change in y over the change in x. The equation for this is (I’ll plug the points in the equation)
$\frac{y_{2}-y_{1}}{x_{2}-x_{1}} = \frac{y_{1}-y_{2}}{x_{1}-x_{2}} = \frac{-1-(-2)}{2-(-2)}= \frac{-1+2}{2+2} =1/4$

Therefore, our answer is the bottom one as our slope is 1/4.

Feel free to ask further questions!

5 0

3 years ago

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Brandon is on one side of a river that is 50 m wide and wants to reach a point 300 m downstream on the opposite side as quickly

AlexFokin [52]

Let P be Brandon's starting point and Q be the point directly across the river from P.
Now let R be the point where Brandon swims to on the opposite shore, and let 
QR = x. Then he will swim a distance of sqrt(50^2 + x^2) meters and then run 
a distance of (300 - x) meters. Since time = distance/speed, the time of travel T is 

T = (1/2)*sqrt(2500 + x^2) + (1/6)*(300 - x). Now differentiate with respect to x: 

dT/dx = (1/4)*(2500 + x^2)^(-1/2) *(2x) - (1/6). Now to find the critical points set 
dT/dx = 0, which will be the case when 

(x/2) / sqrt(2500 + x^2) = 1/6 ----> 

3x = sqrt(2500 + x^2) ----> 

9x^2 = 2500 + x^2 ----> 8x^2 = 2500 ---> x^2 = 625/2 ---> x = (25/2)*sqrt(2) m, 

which is about 17.7 m downstream from Q. 

Now d/dx(dT/dx) = 1250/(2500 + x^2) > 0 for x = 17.7, so by the second derivative 
test the time of travel, T, is minimized at x = (25/2)*sqrt(2) m. So to find the 
minimum travel time just plug this value of x into to equation for T: 

T(x) = (1/2)*sqrt(2500 + x^2) + (1/6)*(300 - x) ----> 

T((25/2)*sqrt(2)) = (1/2)*(sqrt(2500 + (625/2)) + (1/6)*(300 - (25/2)*sqrt(2)) = 73.57 s.



mind blown

8 0

3 years ago