+0

0
273
3

https://gyazo.com/672bc5fcc8315ff551a668355568c028

The task is to decide the x cordinate for the point "P"

The point "P" is 10 "l.e" away from origo

Line A have the equation y=2x-5

l.e stands for a measurement which is not specified

Guest May 23, 2017

#1
+7339
+1

This is what I came up with, though this might not be the intended approach!

I found and labeled the y-intercept and x-intercept of line A, then used that to label the lengths of the blue and pink lines. I labeled the green line "1s" and the orange line "2s" because the slope of line A is 2. So, for every 1 unit line A goes right, it goes up 2 units. " s " is the unknown scale factor.

the x-coordinate of P = blue line + green line

Let's just look at this right triangle:

We can use the Pythagorean theorem to solve for s.

$$(\frac52+1s)^2+(2s)^2=10^2 \\ \frac{25}{4}+5s+5s^2=100 \\ s^2+s+\frac14=20-\frac54+\frac14 \\ (s+\frac12)^2=19 \\ s=\pm\sqrt{19}-\frac12 \\ s=\sqrt{19}-\frac12$$

So...

the x-coordinate of P  $$=\frac52+\sqrt{19}-\frac12 \\~\\ = 2+\sqrt{19}$$

hectictar  May 23, 2017
#1
+7339
+1

This is what I came up with, though this might not be the intended approach!

I found and labeled the y-intercept and x-intercept of line A, then used that to label the lengths of the blue and pink lines. I labeled the green line "1s" and the orange line "2s" because the slope of line A is 2. So, for every 1 unit line A goes right, it goes up 2 units. " s " is the unknown scale factor.

the x-coordinate of P = blue line + green line

Let's just look at this right triangle:

We can use the Pythagorean theorem to solve for s.

$$(\frac52+1s)^2+(2s)^2=10^2 \\ \frac{25}{4}+5s+5s^2=100 \\ s^2+s+\frac14=20-\frac54+\frac14 \\ (s+\frac12)^2=19 \\ s=\pm\sqrt{19}-\frac12 \\ s=\sqrt{19}-\frac12$$

So...

the x-coordinate of P  $$=\frac52+\sqrt{19}-\frac12 \\~\\ = 2+\sqrt{19}$$

hectictar  May 23, 2017
#2
+92826
+1

CPhill  May 23, 2017
edited by CPhill  May 23, 2017
#3
+7339
+1

Thanks CPhill !!!

hectictar  May 23, 2017