Answer:
Explanation:
a)
First I will calculate the equation of the line that goes trough the points
C (0, 0) and (-4, 3).
y= mx + b
y = [(3-0)/ (-4 -0) ] x + 0
y = (-3/4)x →the slope of tis line is (-3/4)
The slope of line l is the negative reciprocal of (-3/4) the slop of line that goes trough points C (0, 0) and (-4, 3) because the lines are ⊥.
Slope of line l is 4/3
b)
The equation of l is
y= (4/3)x + b
for point (x= -4, y = 3) that belongs to the line l, we have
3 = (4/3) (-4) + b
3 = (-16/3) + b
3 + (16/3) = b
(9+16)/3 = b
25/3 = b
The equation of the line l in slope intercept form is :
y = (4/3)x + (25/3)
c)
The radius of circle C is the distance between C (0, 0) and (-4, 3).
d² = (x2-x1)² +(y2-y2)²
d = √(-4-0)² +(3-0)²
d = √16+ 9
d = 5 → the radius of the circle C is 5
d)
Line l intersects the y-axis at (25/3) as we previously found out.
The radius is 5.
...therefore the distance between the line l's y-intercept and the circumference of the circle is
(25/3 ) - 5 = (25 - 5·3)/3 = 10/3