CPhill

As we found in an earlier post here : http://web2.0calc.com/questions/this-is-part-one-of-the-new-problem#r1....the other diagonal = 70

And the area  = (1/2) * product of the diagonals  =

(1/2)(24)(35)  =  420  units^2

3.The "perpendicular bisector" of the line segment $\overline{AB}$ is the line that passes through the midpoint of $\overline{AB}$ and is perpendicular to $\overline{AB}$.  The equation of the perpendicular bisector of the line segment joining the points $(1,2)$ and $(-5,12)$ is $y = mx + b$. Find m + b

The midpoint of AB  is  [ (1 - 5) / 2, (12 + 2) / 2    =  ( - 4/2 14/2)  =  (-2, 7)

And the slope of AB  =  [ 12 - 2] / [ -5 - 1 ]   =  10 / -6   =   - 5/3

So....the perpedicular bisector will have the slope  3/5

And the equation of this bisector  is

y  = (3/5)(x - - 2) + 7

y  = (3/5)x + 6/5 + 7

 y = (3/5)x + 41/5

So.....m + b  =    3/5  + 41/5  =   44/5

2. Find $B - A$ if the graph of $Ax + By = 7$ passes through $(2,1)$ and is parallel to the graph of $2x - 7y = 3$.

The slope of the first line  =  -A/B  =  the slope of the second line  = 2/7

Which implies that   -7A  = 2B  ⇒  B  = (-7/2)A

So

A(2)  + B(1)  = 7

A(2)   +  (-7/2)A  = 7

-(3/2)A  =  7

A  =  -14/3

B =  (-7/2)(-14/3)  =  98/6  =  49/3

So.......the equation of the first line is

(-14/3)x  + (49/3)y  = 7

And B - A  =     49/3 + 14/3      =   63 / 3  =  21

Here's the graph showing this : https://www.desmos.com/calculator/qrhzf2f3op

1. Line $l$ has equation $y = 4x - 7$, and line $m$ with equation $y = ax + b$ is perpendicular to line $l$ at $(2,1)$. What is the $y$-coordinate of the point on $m$ that has $x$-coordinate 6?

The slope of the perpedicular line  is    - (1/4)

So...the equation of  line   m  is

y  = (-1/4)(x -2)  + 1

y  = (-1/4)x + 1/2 + 1

y  = (-1/4)x + 3/2 

So  on this line....when x  = 6

y  = (-1/4)(6) + 3/2

y  = -6/4 + 3/0

y = -3/2  + 3/2

y = 0

If we draw a segment parallel to the base connecting the 1st vertex from the bottom left to the other side....we will have a rectangle  whose width = 24  and its height = 20.....so its area  = 480 units^2

And drawing a segment from this vertex to the 1st vertex from the bottom right will form two triangles....one whose height  = 7 and its base  = 24....so its area  =  (1/2) (7)(24)  = 84 units^2

And a right triangle will also be formed at the top....its area  = (1/2) product of the legs  =

(1/2) * 15 * 20  =  150 units^2

So.....the total area  =  [ 480 + 84 + 150 ]  = 714  units^2

All ides are equal....so......one side is

148/ 4  =   37  units

And the diagonals bisect each other and meet at right angles

And the side forms the hypotenuse of a right triangle and another leg is 1/2 of a diagonal length.....and the remaining leg is 1/2 of the other diagonal length

So....the length of the other diagonal  is

2 * √ [ 37^2 - 12^2 ]  = 2 √1225   =  2 * 35   =    70 units

 $(1,7)$ and $(-2,-3)$ 

Slope  =   [ -3 - 7 ] / [ - 2 - 1]  =  -10 / - 3   =  10 / 3

Equation of the line passing through these points  =

y =  (10/3) (x - 1) + 7

y = (10/3)x - 10/3 + 7

y= (10/3)x  +11/3    multiply through by 3

3y = 10x + 11    rearrange as

10x - 3y  = -11

Interest  earned =

Amount accumulated  - Original amount  = 

10000 ( 1 + .045)^2  - 10000  =  $920.25

I.  direct

II. inverse

III. joint

Let M  =   [ ( a + c) / 2  , (b + d) / 2  ]    = (m, n)

A'  =  (a +14  , b + 20)       B'  =  (c - 2, d - 4)

So  M'  =    [ ( [a + c]  + 12) / 2   ,  ( [b + d]  + 16)/ 2 ]  =   ( m + 6, n + 8)

So..... the distance between M  and M'   is

√ [ (m + 6 - m)^2  + ( n + 8 - n)^2 ]  =  √ [  6^2 + 8^2]  = √ [36 + 64]  = √100  =

10 units