Question

NO LINKS!!! URGENT HELP PLEASE!!!

Find a formula that expresses the fact that an arbitrary point P(x, y) is on the perpendicular bisector "l" of segment AB

A(-3, 3), B(7, -7)

Answer 1

The midpoint of segment AB is: M = ((-3 + 7)/2, (3 + (-7))/2) = (2, -2)

The slope of segment AB is: m = (y2 - y1)/(x2 - x1) = (-7 - 3)/(7 - (-3)) = -1

The slope of a line perpendicular to AB is the negative reciprocal of m: m_perp = -1/m = 1

The equation of the line passing through M with slope m_perp is: y - (-2) = 1(x - 2)

Simplifying, we get: y = x - 4

Therefore, an arbitrary point P(x, y) is on the perpendicular bisector of segment AB if and only if its coordinates satisfy the equation: y = x - 4

Answer 2

y = x - 4

Explanation:

To find the perpendicular bisector of the line segment AB, we first need to find the midpoint of AB. The midpoint M can be found using the midpoint formula:

M = ((x1 + x2)/2, (y1 + y2)/2)

where (x1, y1) = A and (x2, y2) = B

M = ((-3 + 7)/2, (3 - 7)/2) = (2, -2)

The slope of AB can be found using the slope formula:

m = (y2 - y1)/(x2 - x1)

m = (-7 - 3)/(7 - (-3)) = -10/10 = -1

The slope of the perpendicular bisector of AB is the negative reciprocal of the slope of AB. So, the slope of the perpendicular bisector is:

m_perp = 1/m = -1/-1 = 1

Now we have the midpoint M and the slope of the perpendicular bisector. We can use the point-slope form of a line to find the equation of the perpendicular bisector.

y - y1 = m(x - x1)

where m = 1 and (x1, y1) = (2, -2)

y + 2 = 1(x - 2)

y = x - 4

Therefore, the formula that expresses the fact that an arbitrary point P(x, y) is on the perpendicular bisector of segment AB with endpoints A(-3, 3) and B(7, -7) is:

y = x - 4