Locus problem?

Question

Find the equation of the locus of a point that moves so that its distance from the line 3x+4y+5=0 is always 4 units. 

Please show all working out

Answer 1

Finding Locus :

The perpendicular distance formula, this tells us the shortest distance between a point and a line :
D = |ax₁ + by₁ + c| / √(a² + b²)

Where x₁ and y₁ are the coordinates of the point, in this case the point is variable so we will just use x and y which will give us a cartesian equation.

The only difference between this one and the others is that you are given the distance in lieu of a fixed point.

Now the line is 3x + 4y + 5 = 0 and distance (D) = 4

4 = |3x + 4y + 5| / √(3² + 4²)
4√(9 + 16) = |3x + 4y + 5|
4√25 = |3x + 4y + 5|
4(5) = |3x + 4y + 5|
|3x + 4y + 5| = 20

Now solve this absolute value equation, this will yield two cartesian linear equations as expected.

3x + 4y + 5 = 20 and 3x + 4y + 5 = - 20

⇒ 3x + 4y = 15 and 3x + 4y = - 25

Therefore, the equations of the loci are 3x + 4y = 15 and 3x + 4y = - 25.