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 solve the system of equations formed by their equations.

1. The equation of the tangent at point B: -11x + 6y + 64 = 0

2. The equation of line L passing through point A(-3, -2) and the origin (0, 0) with a gradient of 2/3:
y = (2/3)x

To find the point of intersection, substitute the value of y in equation 2 into equation 1:

-11x + 6(2/3)x + 64 = 0

To solve this equation, simplify it:

-11x + 4x + 64 = 0

Combine like terms:

-7x + 64 = 0

Now solve for x:

-7x = -64

Divide both sides by -7:

x = 64/7

Now substitute this value of x back into equation 2 to find the y-coordinate:

y = (2/3)(64/7)

Simplify:

y = 128/21

Therefore, the coordinates where line L and the tangent at point B intersect are (64/7, 128/21).