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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.

User Johanne
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1 Answer

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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}

Let a(-7, 4) and b(5, -12) be points in the plane. (a) find the slope of the line-example-1
Let a(-7, 4) and b(5, -12) be points in the plane. (a) find the slope of the line-example-2
User Jonas Lincoln
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