3

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 $(-2,1)$?

Tim
  • 520
  • 2
  • 6
  • 17
  • There are infinite points on the line. How do you want to decide which points to choose? – Zaros May 18 '16 at 15:06
  • @lordoftheshadows 4 units away from (-2,1). Imagine standing at (-2,1) and walking 4 steps. Where are you now (there are only 2 possible locations)? – Tim May 18 '16 at 15:08

2 Answers2

2

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

Tim
  • 520
  • 2
  • 6
  • 17
  • The right-hand side has to be $4^2$, not $4$. – mathlove May 18 '16 at 15:11
  • @mathlove Thanks, edited. – Tim May 18 '16 at 15:42
  • You made a mistake after you obtained $$(x + 2)^2 + \left(\frac{7}{4}x + \frac{9}{2} - 1\right)^2 = 4^2$$ You should have obtained \begin{align} (x + 2)^2 + \left(\frac{7}{4}x + \frac{7}{2}\right)^2 & = 16\ (x + 2)^2 + \left[\frac{7}{4}(x + 2)\right]^2 & = 16\ (x + 2)^2 + \frac{49}{16}(x + 2)^2 & = 16\ \frac{65}{16}(x + 2)^2 & = 16\ (x + 2)^2 & = \frac{256}{65}\ x + 2 & = \pm \frac{16}{\sqrt{65}}\ x + 2 & = \pm \frac{16\sqrt{65}}{65}\ x & = -2 \pm \frac{16\sqrt{65}}{16}\end{align} – N. F. Taussig May 19 '16 at 09:31
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.

Pratyush Yadav
  • 913
  • 1
  • 6
  • 20