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, we need to solve the system of equations formed by the equation of the tangent and the equation of line L.
Step 1: Write the equation of line L passing through point A(-3, -2) and the origin:
The gradient of line L is given as 2/3. We can use the point-slope form of a linear equation to write the equation:
(y - y1) = m(x - x1)
Substituting the values of the gradient (m = 2/3) and the point A (-3, -2), we have:
(y - (-2)) = (2/3)(x - (-3))
(y + 2) = (2/3)(x + 3)
Expanding and rearranging the equation, we get:
3(y + 2) = 2(x + 3)
3y + 6 = 2x + 6
3y - 2x = 0
So, the equation of line L is 3y - 2x = 0.
Step 2: Write the equation of the tangent at point B(8, 4):
The equation of the tangent is given as -11x + 6y + 64 = 0.
Step 3: Solve the system of equations:
To find the coordinates where the two lines intersect, we need to solve the system of equations:
3y - 2x = 0 (equation of line L)
-11x + 6y + 64 = 0 (equation of the tangent at point B)
There are several ways to solve this system of equations. One method is substitution.
Solving the first equation for y, we get:
3y = 2x
y = (2/3)x
Substituting this value of y in the second equation, we have:
-11x + 6(2/3)x + 64 = 0
-11x + 4x + 64 = 0
-7x + 64 = 0
-7x = -64
x = -64 / -7
x = 64 / 7
Substituting the value of x back into the first equation, we get:
3y - 2(64 / 7) = 0
3y = (2/7)(64)
3y = 128 / 7
y = (128 / 7) / 3
y = 128 / 21
So, the coordinates at which line L and the tangent at point B intersect are approximately (64/7, 128/21).