The coordinates of point S on the line PS with the equation y = -x + 10 are (5, 5), satisfying the given equation.
In the given scenario, there are points P(0,0), Q(3,5), and R(12,2), and a straight line PS with the equation y = -x + 10. The goal is to find the coordinates of point S.
Firstly, let's analyze the given equation y = -x + 10. This equation is in the slope-intercept form y = mx + b, where m is the slope and b is the y-intercept. In this case, the slope m is -1, and the y-intercept b is 10.
Since S lies on the line PS, the coordinates of S must satisfy the equation y = -x + 10. We can substitute the coordinates of S into this equation and solve for S's coordinates.
Let S have coordinates (xs, ys):
ys = -xs + 10
To find xs, set ys equal to 5, as given by the coordinates of point Q:
5 = -xs + 10
Solving for xs:
xs = 5
Now that we have xs, substitute it back into the equation to find ys:
ys = -5 + 10 = 5
Therefore, the coordinates of point S are (5, 5).