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1)Find the equation whose graph is the line passing through the points (2, -3) and 94,5). Enter your answer in standard form Ax + By = C.

 

2)Find the area of the region that is enclosed by the graph of the equation 4x^2 - 12x + 4y^2  + 56y -471 = 0

 

3)The line 3x - 4y +24 intersects the x-axis and y-axis at points A and B respectively. Find distance AB.

 

4)Let  be a point on the graph of equation, (x-94)^2 + (y-49)^2 = 44^2. What is the shortest possible distance between P and the x-axis?

 

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

 

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)".

 

7)Let p and q be constants such that the graph of x^2 + y^2 - 6x +py + q is tangent to the y-axis. What is the area of the region enclosed by the graph?

 

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.

Picture: https://latex.artofproblemsolving.com/1/7/5/1759cfad134302d78a2e2311360b02e6cc81fc65.png

 

9)The circles x^2 + y^2 = 4 and (x-2)^2 + (y-3)^2 = 7 intersect in two points A and B. Find the slope of AB.

 

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.

Picture:https://latex.artofproblemsolving.com/8/d/6/8d69e6ecf5ba0eefdc0b6b3d275445fc2fb678cc.png

 

11)The perimeter of square ABCD is 2. Two cars start from A and B simultaneously and drive clockwise on the perimeter of square ABCD in the same speed. A drone maintains its location as the midpoint of the two cars. How far did the drone fly after both cars get back to where they started?

FiestyGeco  May 16, 2018
 #1
avatar+89775 
+1

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

 

 

 

cool cool cool

CPhill  May 16, 2018
 #2
avatar+89775 
+1

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  

 

 

 

cool cool cool

CPhill  May 16, 2018
 #3
avatar+89775 
+1

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

 

 

 

cool cool cool 

CPhill  May 16, 2018
 #4
avatar+89775 
+1

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

 

 

 

cool cool cool

CPhill  May 16, 2018
 #5
avatar+89775 
+1

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)

 

 

cool cool cool

CPhill  May 16, 2018
 #6
avatar+89775 
+1

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

 

 

cool cool cool

CPhill  May 16, 2018
 #7
avatar+89775 
+1

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

 

 

cool cool cool

CPhill  May 16, 2018
 #8
avatar+89775 
+1

9)The circles x^2 + y^2 = 4 and (x-2)^2 + (y-3)^2 = 7 intersect in two points A and B. Find the slope of AB.

 

 

Simplify the second equation

x^2 - 4x + 4  + y^2 - 6y + 9  = 7

x^2 -4x +  y^2  - 6y  =  -6

 

Add both equations

x^2  - 4x + y^2 - 6y  =  - 6

x^2         + y^2         =    4

______________________

-4x   - 6y  =    -10

 

Solve this for y

4x +  6y  =  10

2x + 3y  = 5

3y  = 5 - 2x

y = [ 5 - 2x ]  / 3       sub this into  x^2 + y^2  = 4  for y

 

x^2  + ( [ 5 -2x ] / 3 )^2  =  4

x^2  +  [ 4x^2 - 20x + 25 ] / 9  = 4

9x^2  + 4x^2 - 20x + 25  = 36

13x^2 - 20x - 11  =  0

 

Solving this equation for  x  produces

 

x  = [10 + 9√3 ]  / 13      or     x  =   [ 10 - 9√3] / 13

 

And  when x   = [ 10 - 9√3] / 13, then y = [15 + 6√3 ] / 13

And  when x  = [10 + 9√3] / 13, then y  = [15 - 6√3] / 13

 

So....the points of intersection are

 

( [ 10 - 9√3 ] / 13, [15 + 6√3 ] / 13 )     and   ( [ 10 + 9√3 ] / 13, [15 - 6√3 ] / 13 ) 

 

And  the slope between these points   is

 

[ ( 15 + 6√3) - ( 15 - 6√3)  ]  /  [ (10 - 9√3)  - (10 + 9√3) ]  =

 

[12√3 ] / [ (-18√3]  =

 

-2/3

 

 

 

  cool cool cool

CPhill  May 16, 2018
 #9
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0

You realize that these are copyrighted problems, right? It literally says "Copyright © AoPS Incorporated. This page is copyrighted material. You can view and print this page for your own use, but you cannot share the contents of this page with others" on the site where these homework problems are.

Guest May 17, 2018
 #10
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Fiestygeco, if you are arrested and imprisoned for this, I’ll bribe the guards and have them smuggle in some juicy bugs for you. June Bugs and May Beetles are my favorite—and they are in season now.

Yours truly,

The Salamander

Guest May 17, 2018

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