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).