Functions (1 Viewer)

[FDownes]

I'm not sure how to tackle this question, can anyone help me out? It asks;

The point P (x, y) is equidistant from the lines y = -2 and 3x - 4y - 12 = 0, and lies in the shaded region of the diagram. Find the equation of the locus of P.

 

[FDownes]

Apparently the answer is 3x - 9y - 22 = 0, x > 4/3, but so far these sheets have been fairly unreliable, so who can tell?

 

[lyounamu]

Apparently the answer is 3x - 9y - 22 = 0, x > 4/3, but so far these sheets have been fairly unreliable, so who can tell?

My bad sorry. I did get that answer. I just didn't work out from my third line. I just put in my incomplete answer. LOL

My working out is like the following:

I worked out the horizontal distance between 3x - 4y - 12 = 0 and (x,y)
And I also worked out the horizontal distance between y=-2 and (x,y)

So I got:

Absolute value of (3x-4y-12)/5 = Absolute value of (y+2)
Then simplifying I get: 3x - 9y - 22 = 0 or 3x + y -2 = 0

But only 3x-9y-22=0 is in the shodow region so the answer is 3x-9y-22=0.

 

[lolokay]

use distance formula

(3x - 4y - 12)/sqrt(3^2 + 4^2) = (y + 2)/(1^2)
3x - 4y - 12 = 5y + 10
3x - 9y - 22 = 0
giving a gradient of 1/3 which lies between the gradients of the other 2 lines

I assume it only lies beyond the point of intersection of the lines;
3x - 4y - 12 = 0
y = -2
3x + 8 - 12 = 0
x = 4/3

so x>4/3

 

[bored of sc]

Would you use perpendicular distance formula?

Damn. Got beaten to it.

 

[lyounamu]

Would you use perpendicular distance formula?

Damn. Got beaten to it.

Yeah. I did.