Question

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

(a) find the slope of the line that contains a and b.

(b) find an equation of the line that passes through a and b. what are the intercepts?

(c) find the midpoint of the segment ab.

(d) find the length of the segment ab.

i

(e) find an equation of the perpendicular bisector of ab.

(f) find an equation of the circle for which ab is a diameter.

Answer 1

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

b) We know the equation is
`y=-(4)/(3)x+b`, where b is a constant. If we substitute in the coordinates of point a, we get that:

`4=-(4)/(3)(-7)+b\\4=(28)/(3)+b\\b=-(16)/(3)`

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

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

`0=-(4)/(3)x-(16)/(3)\\(16)/(3)=-(4)/(3)x\\x=-4`

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

c)
`\left((-7+5)/(2), (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=(3)/(4)(x+1)\\y-4=(3)/(4)x+(3)/(4)\\\boxed{y=(3)/(4)x+(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}`