given O(0,0) and N(12,0) write an equation in terms of x and y for all points P(x,y) such that segment PO is perpendicular to segment PN. simplify this equation and show that P is on a circle.
To find the equation in terms of x and y for all points P(x,y) such that segment PO is perpendicular to segment PN, we can follow these steps:
Step 1: Find the slopes of segments PO and PN.
The slope of segment PO is given by:
mPO = (y - 0) / (x - 0) = y / x
The slope of segment PN is given by:
mPN = (y - 0) / (x - 12) = y / (x - 12)
Step 2: Since perpendicular lines have slopes that are negative reciprocals of each other, we have:
mPO * mPN = -1
Substituting the values we obtained earlier, we can set up the equation:
(y / x) * (y / (x - 12)) = -1
Step 3: Simplify the equation.
Expanding the equation, we have:
(y^2) / (x * (x - 12)) = -1
Cross-multiplying, we get:
y^2 = -x * (x - 12)
Step 4: Rearrange the equation to show that P is on a circle.
Rearranging the equation, we have:
x^2 + y^2 - 12x = 0
This equation represents a circle centered at (6,0) with a radius of 6 units.
Therefore, the simplified equation in terms of x and y for all points P(x, y) such that segment PO is perpendicular to segment PN is:
x^2 + y^2 - 12x = 0, which represents a circle.