Expert: Paul Klarreich - 3/31/2008

Question
I have two questions for you can you help me solve them and then show me how
you solved it.

First Question:
Find

a) the x-intercept A and y-intercept B of the line.

b) the coordinates of the point P so that lin AP is perpendicular to
the line and the magnitude of of the line AP = 1. (There are two answers.)

of the equation (3x)-(4y)=12


Second Question:

Is it possible to find a point on a line that is exactly one point from the x-intercept on a line without using a calculator.  heres a better question,  using the
equation  Y=(-4/3)X+(16/3)find a point on the line that is exactly one unit away
from the x-intercept without using a calculator. (There is only two answers)



If you could do this for me and show me how you did it I would be truly
grateful, I sat and tried these problems for like an hour and I could not figure
nothing out.  Thank you and hope to hear from you soon.

Answer
Questioner:   Cornelius
Category:  Advanced Math
Private:  No
 
Subject:  Pre-Calculus
Question:  I have two questions for you can you help me solve them and then show me how
you solved it.

First Question:
Find

a) the x-intercept A and y-intercept B of the line.

>> If you manipulate the equation into the form:
x     y
--- + --- = 1
a     b

then x = a  and y = b are your intercepts.  Or just use these principles:

x-intercept:  Set y = 0, solve for x.
y-intercept:  Set x = 0, solve for y.

......................................


b) the coordinates of the point P so that line AP is perpendicular to
the line

>> What line? The one below?

and the magnitude of the line AP = 1. (There are two answers.)

of the equation (3x)-(4y)=12

>> Sorry, this makes no sense.  Why don't you reword it and resubmit it.  Check the exact wording in your text and send that.
..................................

Second Question:

Is it possible to find a point on a line that is exactly one point from the x-intercept on a line without using a calculator.  heres a better question,  using the
equation  y = (-4/3)x+(16/3)find a point on the line that is exactly one unit away
from the x-intercept without using a calculator. (There is only two answers)

>> OK, this makes some sense.  Here is what you want:  For p(x,y), and the line  y = (-4/3)x+(16/3).

First find that x-intercept:
Set y = 0, solve for x.

(-4/3)x+(16/3) = 0
(4/3)x = (16/3)

4x = 16
x = 4.
The intercept is (4,0)

Now you want a point (x,y) that is:

ON THE LINE.  That means  y = (-4/3)x+(16/3)
ONE UNIT DISTANCE FROM (4,0). That means  sqrt((x - 4)^2 + y^2) = 1

Now substitute and solve.
 
sqrt((x - 4)^2 + y^2) = 1
sqrt((x - 4)^2 + ((-4/3)x+(16/3))^2) = 1

(x - 4)^2 + ((-4/3)x+(16/3))^2 = 1

x^2 - 8x + 16 + (16/9)x^2 - 128/9 x + 256/9 = 1

Multiply by 9:

9x^2 - 72x + 144 + 16x^2 - 128x + 256 = 9

25x^2 - 200x + 400 = 9
25(x^2 - 8x + 16) = 9
25(x - 4)^2 = 9
(x - 4)^2 = 9/25
x - 4 =  +- 3/5

x = 4 +- 3/5

That's your two solutions for x, then substitute to get your two values of y.