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J 0 P(0.10) Q(3,5) (12 S R(12,2) Diberi persamaan garis lurus PS ialah y = -x + 10. Cari koordinat bagi titik S. Given the equation of the straight line PS is y=-x+10. Find the coordinates of point 5.​

User Ramunas
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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).

User Jeremy Scoggins
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