Given the equation of a line, travel X points along it.

Question

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 $(-2,1)$?

Answer 1

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 $x = -2 \pm16\sqrt{65}$

Answer 2

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.