Explanation:
Question 1
so, we need points or segments of AB that have 12 or less ft distance to point C.
that means we need to draw a circle with center in C and with radius 12 ft, and find the intersections with the line.
the regular equating for a circle is
(x - h)² + (y - k)² = r²
with (h, k) being the center of the circle.
so, our circle is
(x - 9)² + (y + 5)² = 12²
the line is in slope-intercept form
y = ax + b
a being the slope, which is the ratio (y coordinate difference / x coordinate difference) between 2 points on the line.
from (-4. -2) to (3, 5)
x changes by +7 (from -4 to 3).
y changes by +7 (from -2 to 5).
the slope (a) is +7/+7 = 1
b we can see on the graph (0, 2) or we can get it by using the coordinates of one of the points in the almost finished equation :
5 = 1×3 + b
b = 2
our line is therefore
y = x + 2
now we use this in the circle equation
(x - 9)² + ((x + 2) + 5)² = 144
(x - 9)² + (x + 7)² = 144
x² - 18x + 81 + x² + 14x + 49 = 144
2x² - 4x + 130 = 144
2x² - 4x - 14 =0
x² - 2x - 7 = 0
the general solution for such a quadratic equation
ax² + bx + c = 0
is
x = (-b ± sqrt(b² - 4ac))/(2a)
in our case this is
x = (2 ± sqrt((-2)² - 4×1×-7))/(2×1) =
= (2 ± sqrt(4 + 28))/2 =
= (2 ± sqrt(32))/2 =
= (2 ± sqrt(16×2))/2 =
= (2 ± 4×sqrt(2))/2 = 1 ± 2×sqrt(2)
x1 = 1 + 2×sqrt(2) = 3.828427125...
x2 = 1 - 2×sqrt(2) = -1.828427125...
the corresponding y values for these intersection points on the line are
y1 = x1 + 2 = 3 + 2×sqrt(2) = 5.828427125...
y2 = x2 + 2 = 3 - 2×sqrt(2) = 0.171572875...
but (x1, y1) are even beyond David's possibilities for movement, so (3, 5) cuts that segment on the line off.
so, yes, she can hit him between the points
(-1.828427125..., 0.171572875...)
and
(3, 5)