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 intersect, we need to find the point of intersection between these two lines.
First, let's find the equation of line L, which passes through point A(-3, -2) and the origin (0, 0) and has a gradient of 2/3.
The gradient-intercept form of a linear equation is given by y = mx + b, where m is the gradient and b is the y-intercept.
Since the line passes through the origin, the y-intercept (b) is 0. Therefore, the equation of line L can be written as y = (2/3)x.
Now, let's find the equation of the tangent at point B(8, 4). The given equation of the tangent is -11x + 6y + 64 = 0.
To find the point of intersection between the tangent and line L, we need to solve the two equations simultaneously:
-11x + 6y + 64 = 0 ... (Equation 1)
y = (2/3)x ... (Equation 2)
Substituting Equation 2 into Equation 1, we get:
-11x + 6(2/3)x + 64 = 0
-11x + 4x + 64 = 0
-7x + 64 = 0
-7x = -64
x = -64 / -7
x = 64/7
Now, substituting x = 64/7 into Equation 2, we can find y:
y = (2/3)(64/7)
y = 128/21
Therefore, the coordinates where line L and the tangent at point B intersect are (64/7, 128/21).
So, the point of intersection is approximately (9.14, 6.10)