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)

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. :)