Question

The line y=2x - 10 cuts the curve x^2 + y^2=100 at the points A and B. Find the coordinates of the points A and B, coordinates of the mid point of line AB and the equation of the perpendicular bisector of the line AB?

Answer 1

First, let's set up the equations to find the coordinates of points A and B.

Since y = 2x - 10 and x^2 + y^2 = 100, we have:

x^2 + (2x - 10)^2 = 100
x^2 + 4x^2 - 40x + 100 = 100
5x^2 - 40x = 0
5x(x - 8) = 0

So, x = 0 or x = 8.

Plugging x = 0 into y = 2x - 10, we get y = -10. Therefore, point A is (0, -10).

Plugging x = 8 into y = 2x - 10, we get y = 6. Therefore, point B is (8, 6).

The midpoint of AB is calculated as:
Midpoint = ((x_A + x_B)/2, (y_A + y_B)/2)
Midpoint = ((0 + 8)/2, (-10 + 6)/2)
Midpoint = (4, -2)

Now, let's find the equation of the perpendicular bisector of AB.

The slope of AB is (y_B - y_A)/(x_B - x_A) = (6 + 10)/(8 - 0) = 16/8 = 2.

Therefore, the slope of the perpendicular bisector is -1/2 (negative reciprocal of the slope of AB).

Using the midpoint (4, -2) and the slope -1/2, the equation of the perpendicular bisector is given by:
y - y_M = m(x - x_M)
y - (-2) = -1/2(x - 4)
y + 2 = -1/2x + 2
y = -1/2x

Therefore, the equation of the perpendicular bisector of line AB is y = -1/2x.