distance solution

Question

 

 

1.  Find the distance between point P(-2,3) and Q(5,4)   Leave as a simplified radical.

 

2.  Find the cooredinats of the point C, halfway between the points A(5,1) and B(-2,7)

 

3.  What is the equation of the perpendicula bisector of the line between the points (2,2) and (6,6)

 

4.  How do I write an equation of a circle with center at the origin and radius 6?

 

5.  How do I write an equation of the circle with center (-2,-3) and radius 8?

Answer 1

Hi L-

I will roll through the questions in the order posted:

 

1.  The distance can be found using the distance formula (which is derived from the Pythagorean theorem):

  ___________________

\/ (x₁ - x₂)² + (y₁ - y₂)²

 

   ___________________
\/ (-2 - 5)² + (3 - 4)²

 

   ______         _
\/ 49 + 1 =  5\/2

 

2.  This just involves taking the mean of the x and y coordinates.  So, you get

 

(-2 + 5)/2 = 1.5

(3 + 4)/2 = 3.5

 

C = (1.5, 3.5)

 

3.  First, use point slope formula for the equation of the line with those two points:

 

y - y₁ = m(x - x₁)

m = (y₁ - y₂)/(x₁ - x₂)

m = (6 - 2)/(6 - 2) = 1

 

y - 2 = 1(x - 2)

y = x

 

Now all we do is find the midpoint of the line to know where the perpendicular bisector hits.  As we saw above, we can calculate the midpoint to be (4, 4).  Now, this line, since its perpendicular, will have a negative inverse slope of the original line.  Therefore, it will be -1.  So, the equation for the line is:

 

y - 4 = -1(x - 4)

y = -x + 4 + 4

y = -x + 8

 

4.  The equation for a circle is as follows:

 

(x-a)² + (y-b)² = r² where a and b are the coordinates of the center, and r is the radius.

 

So, you get:

 

(x-0)² + (y-0)² = 6²

x² + y² = 36

 

5.  Using the equation above:

(x+2)² + (y+3)² = 64

 

 

Let me know if you have any questions.