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Felix2008 · Mar 31, 2008, 01:36 PM

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 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. Here's 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)

galactus · Mar 31, 2008, 04:12 PM

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.Yes, it is here's 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)

$y=\frac{-4}{3}x+\frac{16}{3}$

We can use the distance formula.

We can easily find the x-intercept by setting y=0 and solving for x.

We find quickly that x=4.

Using the coordinates of the x-intercept which are (4,0).

$(x-4)^{2}+(y-0)^{2}=1$

$(x-4)^{2}+y^{2}=1$

But, $y=\frac{-4}{3}x+\frac{16}{3}$ , sub that in:

$(x-4)^{2}+(\frac{-4}{3}x+\frac{16}{3})^{2}=1$

$\frac{25}{9}x^{2}-\frac{200}{9}x+\frac{400}{9}=1$

Multiply by 9 to get rid of the fractions:

$25x^{2}-200x+400=9$

$25x^{2}-200x+391=0$

Factor:

$(5x-23)(5x-17)=0$

Now, it's easy to find the two solutions. Plug those back into the line equation to find their corresponding y values. Those are the two points which are 1 unit along the line from the x-intercept. And no calculator. :)