Another Locus Question (1 Viewer)

[Pinchy444]

Hi,

I am struggling with yet another Locus question; , any help will be appreciated.

The answer is; 3x+4y+25=0 , 3x+4y-15=0

Thanks!

 

[deswa1]

Make the point an arbitrary x1,y1 and use the perpendicular distance formula.

 

[Pinchy444]

Could you please show working?

 

[Timske]

<a href="http://www.codecogs.com/eqnedit.php?latex=\frac{ax_{1} @plus; by_{1}@plus; c}{\sqrt{a^2@plus;b^2}} = 4 \\\\ \frac{3x_{1} @plus; 4y_{1} @plus; 5}{\sqrt{25}}= 4 \\\\ 3x @plus; 4y @plus; 5 = \pm 20 \\\\ 3x @plus; 4y @plus; 25 = 0~ and~ 3x @plus; 4y - 15 = 0" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{ax_{1} + by_{1}+ c}{\sqrt{a^2+b^2}} = 4 \\\\ \frac{3x_{1} + 4y_{1} + 5}{\sqrt{25}}= 4 \\\\ 3x + 4y + 5 = \pm 20 \\\\ 3x + 4y + 25 = 0~ and~ 3x + 4y - 15 = 0" title="\frac{ax_{1} + by_{1}+ c}{\sqrt{a^2+b^2}} = 4 \\\\ \frac{3x_{1} + 4y_{1} + 5}{\sqrt{25}}= 4 \\\\ 3x + 4y + 5 = \pm 20 \\\\ 3x + 4y + 25 = 0~ and~ 3x + 4y - 15 = 0" /></a>

 

[deswa1]

Could you please show working?

Yep but I'm doing something else atm. If no one does it in about an hour, I'll do it with working.

EDIT: Clearly you didn't have to wait an hour lol

 

[Pinchy444]

LOL, yer guess I didn't!

Appreciate the help from both of you.

Don't be surprised if I put another question up tomorrow..

 

[deswa1]

LOL, yer guess I didn't!

Appreciate the help from both of you.

Don't be surprised if I put another question up tomorrow..

Go for it- post any question and someone will answer it.