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, follow these steps:
Step 1: Find the equation of line L passing through point A(-3, -2) and the origin (0, 0) with a gradient of 2/3.
The equation of a line in slope-intercept form 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 (c) is 0. Therefore, the equation of line L is y = (2/3)x.
Step 2: Find the equation of the tangent line at point B(8, 4).
The given equation of the tangent is -11x + 6y + 64 = 0. To convert it to slope-intercept form, isolate y:
-11x + 6y + 64 = 0
6y = 11x - 64
y = (11/6)x - (64/6)
y = (11/6)x - 32/3
Step 3: Set the equations of line L and the tangent line equal to each other and solve the system of equations to find the intersection point.
Substitute the equation of line L (y = (2/3)x) into the y-value of the tangent equation:
(11/6)x - 32/3 = (2/3)x
Multiply through by 6 to eliminate the denominators:
11x - 64 = 4x
Simplify and isolate x:
11x - 4x = 64
7x = 64
x = 64/7
Substitute this x-value back into the equation of line L to find the y-value:
y = (2/3)(64/7)
y = 128/21
So, the coordinates where the line L and the tangent at point B intersect are approximately (64/7, 128/21).