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1)Find the distance between (3,4) and the line 4x + 3y + 7 = 0.

 

2)Find the distance between the graphs of 5x - 12y + 45 = 0 and 5x - 12y - 46 = 0.

 

3)A circle centered at the origin is tangent to the line x - y + 4 = 0. What is the area of the circle?

 

4)Find the area of the region bounded by the lines y = 5 and 5x + 2y = 25 and the coordinate axes.

 

5)We are given A = (2,5), B = (-5, 2), C = (-2, -5), and D = (5,-2). Find [ABCD].

 

6)For some positive real number r, the line x + y = r is tangent to the circle x^2 + y^2 = r. Find r.

 

7)We are given O = (0,0), A = (5,2) and B = (6, -3). Find [OAB].

 

8)A circle centered at (3,2) is tangent to the line y = (x+1)/3. Find the area of the circle.

 

9)A line with slope 3 is 2 units away from the origin. Find the area of the triangle formed by this line and the coordinate axes.

 

10)The line y = (3x + 20)/4 intersects a circle centered at the origin at A and B. We know the length of chord AB is 20. Find the area of the circle.

 

11)The lines y = (5/12)x and y = (4/3)x are drawn in the coordinate plane. Find the slope of the line that bisects the acute angle between these lines.

Picture: https://latex.artofproblemsolving.com/4/b/c/4bc61e6a0f1f043a88e9efa268018820d89cea3a.png

FiestyGeco  May 17, 2018
 #1
avatar+91071 
+1

1)Find the distance between (3,4) and the line 4x + 3y + 7 = 0

 

We can use this to find the answer.....the distance between  (a, b)  and the line  AX + BY + C  = 0  is given by

 

l A(a)  +  B(b)  +  C  l   

_________________

  √ [ A^2 + B^2 ]

 

So we have

 

  l  4(3) + 3(4)  l         l  24  l               24

  ___________  =     _______  =      ___  =    4.8 units

  √ [ 4^2 + 3^2]           √25                  5

 

 

cool cool cool

CPhill  May 18, 2018
 #2
avatar+91071 
+1

2)Find the distance between the graphs of 5x - 12y + 45 = 0 and 5x - 12y - 46 = 0.

 

The first line has a y intercept  of 

-12 y = -45

y = 45 / 12

y = 15/4  = 3.75

 

So....the point  (0, 3.75)  is on the first line

 

And the distance betwen this point and the second line will be the perpendicular distance between the two lines....so...we have

 

l 5 (0)  - 12 (3.75) - 46  l             l 91 l                    91

____________________  =    ___________  =   ___   =   7   units

   √[ 5^2 + 12^2]                         √169                    13

 

 

 

cool cool cool

CPhill  May 18, 2018
 #3
avatar+91071 
+1

3)A circle centered at the origin is tangent to the line x - y + 4 = 0. What is the area of the circle?

 

The distance from the centrer to the line will be the radius of the circle....so we have

 

l (0)  - (0) +  4  l                        4

______________  =              ____   =  2√2  =  √8  =  the radius 

   √[ 1^2 + 1^2]                        √2

 

So...the area of the circle  is

 

pi * (  √8)^2   =

 

8pi   units^2

 

 

 

 

cool cool cool

CPhill  May 18, 2018
 #4
avatar+91071 
+1

4)Find the area of the region bounded by the lines y = 5 and 5x + 2y = 25 and the coordinate axes.

 

Look at the  following pic:

 

 

This are will be the difference  in the area  of triangle ADB  - area of triangle ACE

 

The area of triangle  ADB  =  (1/2) base (height)  = (1/2) (5)(12.5)  =  31.25 units^2

 

Since the triangles are similiar, the linear scale factor of ACE to ADB  = 3/5

 

So...the area of triangle  ACE  =  area of ADB * (3/5)^2  = 31.25 * (3/5)^2  = 11.25 units^2

 

So...the area of the bounded region is   31.25 - 11.25 =   20 units^2

 

 

cool cool cool

CPhill  May 18, 2018
 #5
avatar+91071 
+1

5)We are given A = (2,5), B = (-5, 2), C = (-2, -5), and D = (5,-2). Find [ABCD].

 

We can split this up  into  the area of two triangles.....ABD  and CBD

 

BD will serve as the base for both

And its length is  √ [ (-5-5)^2  + (-2 -2)^2 ] =√[ 10^2 + 4^2  ]  = √116 =  2√29 units

 

And the slope of BD  is  [ -2 - 2 ] / [ 5 - -5]  =  [-4] / [10]  = -2/5

And the equation of the line containing  BD  is

 

y = (-2/5)(x -5) - 2

5y = -2(x - 5) - 10

5y  = -2x + 10 - 10

5y + 2x = 0

2x + 5y  = 0

 

So...the distance from A to this line will be the height of triangle ABD  =

 

l 2(2) + 5(5) l                     29

___________   =           ____  =  √29  units

√ [2^2 + 5^2 ]                  √29

 

So...the area  of  triangle ADB  = (1/2) *2√29 *√29  =  29   units^2

 

 

And the distance  between C and this line will be the height of triangle CBD

 

l 2(-2)  + 5(-5)  l              29

_____________  =       _____  =  √29  units

 √29                                √29

 

And  triangle CBD will have the same area as triangle ABD

 

So...the total area  of  [ABCD ] =    58 units^2

 

 

 

cool cool cool

CPhill  May 18, 2018
 #6
avatar+91071 
+1

6)For some positive real number r, the line x + y = r is tangent to the circle x^2 + y^2 = r. Find r.

 

At the point of tangency,  x  = x^2  and  y  = y^2

And these will be true  at   0  and 1

 

But...x, y  cannot  = 0 since the center of the circle is (0,0)

 

So...x, y  = 1

 

And

 

x + y  =

 

1 + 1  =

 

2   = r

 

 

cool cool cool

CPhill  May 18, 2018
 #7
avatar+91071 
+1

7)We are given O = (0,0), A = (5,2) and B = (6, -3). Find [OAB].

 

We have triangle  OAB

 

Let AB   be the base

 

And its length is

 

√ [ (6 - 5)^2  + (-3 - 2)^2  ]  =  √ [ 1^2  + 5^2  ]  = √26

 

And the slope between these two points  is    [-3 -2 ] / [ 6 -5]  = --5

 

And the equation of the line containing AB  is

 

y  =-5(x - 5) + 2

y  = -5x + 25  + 2

y = -5x + 27

5x + 1y -27  = 0

 

And the distance from (0,0) to this line forms the height of OAB

 

l 5(0)  + 1(0)  -27]                 27

_______________  =        ____

√[ 5^2  + 1^2 ]                    √26

 

 

So....the area  of  OAB   is  (1/2)√26 * 27/√26  =   13.5 units^2

 

 

cool cool cool

CPhill  May 18, 2018
 #8
avatar+91071 
0

I worked  (9)   but my answer disappeared...it is a little lengthy to re-create...but....the answer ends up as  20/3  units^2

 

 

cool cool cool

CPhill  May 18, 2018
 #9
avatar+91071 
+1

8)A circle centered at (3,2) is tangent to the line y = (x+1)/3. Find the area of the circle.

 

y  = (x + 1)  / 3      can be written as

 

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

 

3y = x + 1

 

x - 3y + 1   = 0

 

The distance from ( 3,2) to this line will be the radius of the circle

 

l 3 - 3(2) + 1 l               2             √10          √2

____________  =     ____   =      ___  =     ___

√[ 1^2 + 3^2 ]             √10              5            √5

 

 

So...the area  of the  circle is

 

pi * ( √ (2/5) ) ^2  =

 

2pi / 5   units^2

 

 

cool cool cool

CPhill  May 18, 2018
 #10
avatar+91071 
+1

10)The line y = (3x + 20)/4 intersects a circle centered at the origin at A and B. We know the length of chord AB is 20. Find the area of the circle.

 

We can write the line as

 

4y  = 3x + 20

 

3x  - 4y  + 20  = 0

 

The perpendicular  distance between the center of the circle and the line will bisect the chord

And this distance is

 

l  3(0) - 4(0)  + 20  l            20

________________  =   ______ =     4

    √[3^2 + 4^2] =                 5

 

A right triangle will be formed by the bisected chord, the perpendicular distance from the center to the midpoint of the chord and the radius which will form the hypotenuse.

 

So....the radius  is √[10^2 + 4^2 ]  =  √116  units

 

And the area  of the circle is

 

pi *  ( √116)^2  =

 

116 pi   units^2

 

 

 

cool cool cool

CPhill  May 18, 2018
 #11
avatar
+1

Why in the world are you posting copyrighted problems? It's illegal, and it also deprives the whole point of the Geometry class. The whole point of the homework is to practice the problems so you can study for the midterm, and asking others to do it for you won't help at all. Besides, the AoPS Admins already know you are cheating, so doing this won't help at all.

Guest May 18, 2018
 #12
avatar+91071 
+1

11)The lines y = (5/12)x and y = (4/3)x are drawn in the coordinate plane. Find the slope of the line that bisects the acute angle between these lines.

 

Construct a circle centered at the origin with a radius of 1

 

The  equation  is

 

x^2 + y^2  = 1

 

We can find the x coordinate  of the intersection of the first line and this circle,thusly "

 

x^2  + [ (5/12)x ]^2  = 1

x^2 + (25/144)x^2   =1

[144 + 25] / 144  x^2  = 1

[169] /144 x^2  =1

x^2 = 144/169

x = 12/13

And y = (5/12)(12/13) = 5/13

So   ( 12/13, 5/13 )  is on the circle

 

Like wise...we can find the intersection of this circle with the  second line :

 

x^2  +[ (4/3)x\^2  =1

x^2  + (16/9)x^2  = 1

[9 + 16 ] / 9 * x^2  =1

25/9 x^2  =1

x^2  = 9/25

x = 3/5

And y = (4/3)(3/5)  = 4/5

So  (3/5, 4/5)  is on the circle

 

And a chord can be drawn on this circle with endpoints   ( 12/13, 5/13)  and (3/5, 4/5 )

And the midpoint of this chord is

  (  [12/13 + 3/5]  / 2,  [ 5/13 + 4/5 ] / 2 )  =

 

(99/130 , 77/130  )

 

And  the line drawn  from the origin to this point is the line we seek and it will have the slope

 

[77/130]  / [99/130]  =    77/99   =  7/9

 

 

cool cool cool

CPhill  May 18, 2018

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