A(-3,-2) B(8,4)are the ends of the diameter of a circle.
the tangent of the circle at point B is -11x+6y+64=0
the line L passes through A and the origin. line L has a gradient of 2/3
what is the coordinates that the line L and tangent at point B cross??
(this question had 4 parts, ive done the first 3 and i think ive got the calculations right. im just stuck on the 4th part)
please help thanks
To find the coordinates where line L and the tangent at point B cross, you need to set up and solve a system of equations.
Step 1: Equations of the lines:
The equation of line L passing through point A (-3, -2) and the origin (0, 0) with a gradient of 2/3 can be found using the point-slope form of a line:
y - y1 = m(x - x1)
where (x1, y1) is a point on the line and m is the slope.
Using the point A (-3, -2):
y - (-2) = (2/3)(x - (-3))
y + 2 = (2/3)(x + 3)
Simplifying the equation:
3y + 6 = 2x + 6
2x - 3y = 0 ... (Equation 1)
The equation of the tangent at point B (8, 4) is given as:
-11x + 6y + 64 = 0 ... (Equation 2)
Step 2: Solving the system of equations:
To find the coordinates where the two lines cross, solve the system of equations formed by Equation 1 and Equation 2 simultaneously.
To do this, you can use one of several methods, such as substitution or elimination. Here, we will use the method of substitution:
Starting with Equation 1:
2x - 3y = 0
Rearranging the equation to solve for x:
x = (3/2)y ... (Equation 3)
Substitute Equation 3 into Equation 2:
-11((3/2)y) + 6y + 64 = 0
Solving for y:
-33/2y + 6y + 64 = 0
(-33/2 + 12/2)y + 64 = 0
(-33 + 12)y + 64 = 0
(-21)y + 64 = 0
-21y + 64 = 0
-21y = -64
y = 64/21
Substituting y = 64/21 back into Equation 3 to find x:
x = (3/2) * (64/21)
x = 3 * 64 / (2 * 21)
x = 192/42
x = 32/7
The coordinates where line L and the tangent at point B cross are (32/7, 64/21).