Question

find the equation of the locus of a point which moves so that its distance from the line 6x+2y=13 is always twice its distance from the line y=3x+8

Answer 1

@freckles

Answer 2

@jim_thompson5910

Answer 3

Do you know how to find the distance from a point to a line?

Answer 4

use sq.rt(x^2+y^3)

Answer 5

?

Answer 6

hmm no not quite

Answer 7

this page

https://en.wikipedia.org/wiki/Distance_from_a_point_to_a_line

offers a great simple formula for finding the distance

https://upload.wikimedia.org/math/2/a/e/2ae89917912e4c37c8673d56ac84fa81.png

Answer 8

keep in mind that they write the standard form equation as `ax+by+c = 0`
so you'll have to turn `6x+2y=13` into `6x+2y-13=0`

Answer 9

ok and 3x-y+8=0

Answer 10

For some reason, the image in the second link is very small

It should say this

\[\Large \text{distance}(ax+by+c=0,(x_0,y_0)) = \frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}\]

Answer 11

Good, `y=3x+8` would turn into `3x-y+8 = 0`

Answer 12

Let's make P be some unknown point (x,y)

Answer 13

The distance from P to the line L1 `6x+2y-13=0` is

\[\Large d(P,L1) = \frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}\]
\[\Large d(P,L1) = \frac{|6x+2y-13|}{\sqrt{6^2+2^2}}\]
\[\Large d(P,L1) = \frac{|6x+2y-13|}{\sqrt{40}}\]

Answer 14

The distance from P to the line L2 `3x-y+8 = 0` is

\[\Large d(P,L2) = \frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}\]
\[\Large d(P,L2) = \frac{|3x-y+8|}{\sqrt{3^2+(-1)^2}}\]
\[\Large d(P,L2) = \frac{|3x-y+8|}{\sqrt{10}}\]

Answer 15

`find the equation of the locus of a point which moves so that its distance from the line 6x+2y=13 is always twice its distance from the line y=3x+8`

This means

\[\Large d(P,L1) = 2*d(P,L2)\]

Answer 16

\[\Large d(P,L1) = 2*d(P,L2)\]
\[\Large \Large \frac{|6x+2y-13|}{\sqrt{40}} = 2*\frac{|3x-y+8|}{\sqrt{10}}\]
\[\Large \Large \left(\frac{|6x+2y-13|}{\sqrt{40}}\right)^2 = \left(2*\frac{|3x-y+8|}{\sqrt{10}}\right)^2\]
\[\Large \Large \frac{(6x+2y-13)^2}{40}= 4*\frac{(3x-y+8)^2}{10}\]

Answer 17

\[(6x+2y-13)^2=16(3x-y+8)^2\]

Answer 18

\[(6x+2y-13)(6x+2y-13)=16(3x-y+8)(3x-y+8)\]\[36x^2+24xy+4y^2-156x-52y+169=16(9x^2-6xy+48x+y^2-16y+64)\]

Answer 19

@jim_thompson5910