Another Locus Question

**Pinchy444** · 7 Jul 2012, 9:05 PM

Hi,

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

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

Thanks!

**deswa1** · 7 Jul 2012, 9:08 PM

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

**Pinchy444** · 7 Jul 2012, 9:16 PM

Could you please show working?

**Timske** · 7 Jul 2012, 9:29 PM

$\frac{ax_{1} + by_{1}+ c}{\sqrt{a^2+b^2}} = 4 \\\\ \frac{3x_{1} + 4y_{1} + 5}{\sqrt{25}}= 4 \\\\ 3x + 4y + 5 = \pm 20 \\\\ 3x + 4y + 25 = 0~ and~ 3x + 4y - 15 = 0$

**deswa1** · 7 Jul 2012, 9:30 PM

Originally Posted by Pinchy444

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