CPhill

10)A circle is tangent to the y-axis at the point (0,2) and passes through the point (8,0) as shown. Find the radius of the circle.

Let  the center of the circle  be  ( a, 2)

So.....the distance from the center of the circle to both points is the same....and we have

( a - 0)^2  + (2 - 2)^2  = ( 8 - a)^2  + (2 - 0)^2

a^2   =   a^2  - 16a + 64  + 4  = 0

-16a + 68  = 0

-16a  = -68

a = 68/16 =  17/4

So...the center  is   (17/4, 2)

And the radius  is   17/4  = 4.25 units

8)In rectangle ABCD, E is the midpoint of BC and F is the midpoint of CD. Let G be the intersection of AE and BF. We know AB = 10 and BC = 15. Find DG.

Let  A  = (0,0)   B = (0, 10)  C  = (15,10)  D  =(15,0)  E  = ( 7.5, 10)  F = (15, 5)

The equation of the the line containing segment AE  is

y = (4/3)x

The slope of segment BF   is  [5 - 10] / [15 - 0] = -5/15  = -1/3

And the equation of the line containing BE  is

y = (-1/3)x + 10

So....the x coordinate of the intersection  of these two lines can be found as

(4/3)x  = (-1/3)x + 10

(4/3 + 1/3) x  = 10

(5/3)x  =10

x = 6

And the y coordinate of the intersection of these lines is

y = (4/3)(6)  = 8

So....G  = ( 6,8)

So  DG   =  √ [ (15 - 6)^2 + (8 - 0)^2 ] = √[9^2 + 8^2 ] = √[81 + 64] = √145 units

6)Given  A = (1,1), B = (4,2), C = (3,4)  and D = (2,3). Find the intersection of the diagonals of quadrilateral ABCD. Enter your answer in the form "(x,y)".

Probably several ways to do this, but let's get the equation of two lines and find their intersection

The line containing AC  has a slope  of   [ 4 - 1 ] / [ 3 - 1 ] = 3/2

And the equation of this line is

y = (3/2) ( x  - 1)  + 1

y = (3/2)x +=- 1/2

The line containing BD  has a slope  of [ 3 - 2] / [ 2 - 4]   = -1/2

And the equation of this line is

y = (-1/2)(x - 2) + 3

y = (-1/2)x + 4

To find the x intersection of these lines, we have

(3/2)x - 1/2  = (-1/2)x + 4

2x  =  9/2

x = 9/4

And the  y coordinate of the intersection is

y = (3/2)(9/4) - 1/2

y = 27/8 - 1/2

y= 27/8 - 4/8

y  = 23/8

So...the intersection of the diagonals  is  ( 9/4, 23/8)

5)The graph x^2 + y^2 = 100 intersects y = 3x at two points P and Q. Find PQ.

Sub the second function into the first function for y

x^2  +  ( 3x)^2   = 100

x^2  + 9x^2  =100

10x^2  = 100     divide both sides  

x^2  = 10       take both roots

x = √10     or    x  =  -√10

So....let P   =  (√10, 3√10)   and  Q  = ( -√10 ,  -3√10 )

So

PQ   =  √ [ (2√10)^2  + (6√10)^2  ]  =  √ ( 40 + 360)  = √400  =  20 units

3)I'm assuming that we have.... 3x - 4y =  24 intersects the x-axis and y-axis at points A and B respectively. Find distance AB.

x intercept  is  (8,0)   

y intercept  is  ( 0, -6)

The distance from A to B  is

√ [ ( 8 - 0)^2  + ( -6 - 0)^2  ]  =

√ [ 64  + 36 ] =

√100  =

10 units

2)Find the area of the region that is enclosed by the graph of the equation

4x^2 - 12x + 4y^2  + 56y -471 = 0      completre the square on x and y

4(x^2 - 3x  + 9/4)  +  4 (y + 14y + 49)  = 471  + 9  + 196

4 ( x - 3/2)^2 + 4 ( y + 7)^2  =  676

(x - 3/2)^2  + (y + 7)^2  =  169

This is a circle  centered  at  (3/2, -7)  with a radius of 13

The area   is      pi * r^2  =    169 pi units^2  

1)Find the equation whose graph is the line passing through the points (2, -3) and (4,5). Enter your answer in standard form Ax + By = C.

Slope   =   [ 5  -  -3  ]  /  [ 4 - 2 ]  =   8/ 2  = 4

Equation is

y  = 4 (x -4) + 5

y  = 4x - 16  + 5

y = 4x   - 11

4x  - 1y  =  11

Independent

P ( A l B)  = P (A)

P (two white cards)  = P(white on first draw) * P (white on second draw)

14 / 95   =   2 / 5   *   P (white on second draw)

Multiply both sides by 5/2

(5/2)(14/95)   = P (white on second draw)   = 70 / 190   =   7 / 19

P  (odd  then even )   =

(5/10)  (4/9)   = 

20 / 90 =

2 / 9