Question

Let a(-7, 4) and b(5, -12) be points in the plane.

Answer 1

a) $m=\frac{-12-4}{5-(-7)}=\frac{-16}{12}=\boxed{-\frac{4}{3}}$

b) We know the equation is $y=-\frac{4}{3}x+b$, where b is a constant. If we substitute in the coordinates of point a, we get that:

$4=-\frac{4}{3}(-7)+b\\4=\frac{28}{3}+b\\b=-\frac{16}{3}$

This means the y-intercept is $\boxed{\left(0, -\frac{16}{3} \right)}$

To find the x-intercept, we can set the equation equal to 0 and solve for x.

$0=-\frac{4}{3}x-\frac{16}{3}\\\frac{16}{3}=-\frac{4}{3}x\\x=-4$

So, the x-intercept is $\boxed{(-4, 0)}$

c) $\left(\frac{-7+5}{2}, \frac{4-12}{2} \right)=\boxed{(-1, -4)}$

d) $ab=\sqrt{(-7-5)^{2}+(-12-4)^{2}}=\boxed{20}$

e) The perpendicular bisector must pass through the midpoint, (-1, 4), and have a slope that is the negative reciprocal (so in this case, 3/4). Substituting into point-slope form,

$y-4=\frac{3}{4}(x+1)\\y-4=\frac{3}{4}x+\frac{3}{4}\\\boxed{y=\frac{3}{4}x+\frac{19}{4}}$

f) If ab is a diameter of said circle, this means it is centered at the midpoint, (-1, 4), and has a radius of 20/2=10. This means the equation is

$\boxed{(x+1)^{2}+(y-4)^{2}=100}$