Given the equation of a line, travel X points along it. I have the equation of a line:
$$y=\frac{7}{4}x\ +\frac{9}{2}$$
I want to find the point 4 units along the line, starting from the point $(-2,1)$.
I tried using Pythagoras and forming this equation:
$$\sqrt{(7a)^2 + (4b)^2} = 4$$
with the gradient - and of course, $a$ and $b$ can be decimals. But there are two unknowns, and the equation I formed was an elipse. I assume it has the solutions in it, but I do not know how to extract them.
How would I find the point 4 away from $(-2,1)$?
 A: Having posted this, I realised I was just asking for the intersection between the line 
$$y=\frac{7}{4}x\ +\frac{9}{2}$$
and the circle around  $(-2,1)$ with radius $4$
$$\left(x+2\right)^2+\left(y-1\right)^{2\ }=\ 4^2$$
Then substitute in y= $y=\frac{7}{4}x\ +\frac{9}{2}$ for 
$$(x+2)^2+({7 \over 4}x\ + {9 \over 2}-1)^{2}=4^2$$
so $x = -2 \pm16\sqrt{65}$
A: You can also use the parametric equation of the line to get your points.
The parametric equation of a line is
$$x = x_1 + r \cos \theta \\
y = y_1 +r\sin \theta$$
Here $x$ and $y$ are the coordinates of a point on the line at a distance of $r$ from the point $(x_1, y_1)$(which should lie on the line). $\theta$ is the angle the line makes with the positive direction of $X$-axis.
Now you just have to fill $x_1 = -2$, $y_1 = 1$ and $r = \pm 4$.
$\cos \theta$ and $\sin\theta$ can be found through the value of $\tan\theta$, which is $\frac 7 4$. We get $\cos\theta = \frac{4}{\sqrt{65}}$ and $\sin\theta = \frac{7}{\sqrt {65}}$.
So, you get $x = -2 + 4\times \frac{4}{\sqrt {65}}$ or $x = -2 - 4\times \frac{4}{\sqrt {65}}$
This gets you $x = -2  - \frac{16}{\sqrt {65}}$.
My answer is a little different from yours. A quick check tells me that $(-2-\frac{16}{\sqrt{65}}, 1-\frac{28}{\sqrt{65}})$ does lie on the line and is at a distance of 4 from $(-2,1)$. Maybe you made a mistake.
