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 the line L and the tangent at point B cross, we can start by solving the system of equations formed by the line L and the equation of the tangent.

First, let's find the equation of the line L passing through point A(-3, -2) and the origin (0, 0) with a gradient of 2/3.

The gradient-intercept form of a linear equation is given by y = mx + c, where m is the gradient and c is the y-intercept.

Since the line passes through the origin (0, 0), the y-intercept is 0.

So, the equation of line L can be written as y = (2/3)x.

Next, let's find the coordinates where the tangent at point B(8, 4) intersects with line L.

We need to find the coordinates (x, y) that satisfy both the equation of the tangent and the equation of line L.

Substituting y = (2/3)x into the equation of the tangent, we get:

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

Simplifying the equation:

-11x + 4x + 64 = 0

-7x + 64 = 0

-7x = -64

x = (-64)/(-7)

x = 64/7

Now, substitute x = 64/7 into the equation of line L to find y:

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

y = (128/3)/7

y = 128/21

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