Pls Help

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

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

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

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

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

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

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

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

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

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

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.

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