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 intersect, you need to solve the system of equations formed by the equation of the tangent line and the equation of line L. Here's how you can do it:
1. Equation of the tangent line:
The given equation of the tangent line is -11x + 6y + 64 = 0. This equation can be rearranged into the slope-intercept form (y = mx + b) by isolating y:
6y = 11x - 64
y = (11/6)x - (64/6)
y = (11/6)x - (32/3)
2. Equation of line L:
Given that line L passes through point A(-3, -2) and the origin (0, 0), we can find its equation using the point-slope form (y - y1 = m(x - x1)), where (x1, y1) are the coordinates of the given point:
y - (-2) = (2/3)(x - (-3))
y + 2 = (2/3)(x + 3)
y + 2 = (2/3)x + 2
y = (2/3)x
3. Solving the system of equations:
Now, we have two equations:
Equation 1: y = (11/6)x - (32/3) (equation of the tangent line)
Equation 2: y = (2/3)x (equation of line L)
To find the coordinates where the two lines intersect, set the right-hand sides of the two equations equal to each other and solve for x:
(11/6)x - (32/3) = (2/3)x
Multiply through by 6 to get rid of the denominators:
11x - 64 = 4x
Move the terms involving x to the left side:
11x - 4x = 64
7x = 64
Solve for x:
x = 64/7
Substitute this value back into either equation to find y:
y = (2/3)(64/7)
y = 128/21
Therefore, the coordinates where line L and the tangent at point B intersect are approximately (64/7, 128/21).