# I have no real clue how to solve this question.. is it with system of equations? But how do I do that and get a line equation answer? Thanks!

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For two non-parallel lines the locus of points equidistant from them are

the two bisectors of angles formed by these lines. Please see the picture bis.png. The initial line are black and bisectors are red and blue.

I can prove this but probably you should know.

For the lines in question we can easily determine the bisectors:

**6**. The point of intersection of y=x+3 and y=-x+3 is (0, 3).

The correct answer must have two lines and the point (0, 3) must be at them. So the only possible correct answer is **(3) y=3 and x=0**.

And this answer is correct because initial lines has slopes 1 and -1 and therefore the angles +-45 degrees to the axes (and the lines from answer are parallel to x-axis and y-axis, respectively).

Please look at the picture bis6.png.

**7**. The given lines y=3 and x=2 intersect at (2, 3), so the bisectors also must intersect here. Futher, these lines are parallel to the axes so bisectors must have an angle +-45 degrees to the axes so the slopes of +-1.

The equation of the first bisector is

y-3 = 1*(x-2) (3 and 2 are from the point of intersection), or**y=x+1**.

The equation of the second bisector is

y-3 = (-1)*(x-2), or**y=-x+5**.

Please ask if something should be explained in more detail.

**Images:**

There is a purely algebraic way to solve this problem, though using symmetry or geometrical arguments are more elegant.

Take a general point in the plane (x,y). To determine the distance from this point to a line given in standard form: point (m,n) and Ax+By+C=0 the distance is:

`d=|Am+Bn+C|/sqrt(A^2+B^2) `

For y=x+3 we rewrite as x-y+3=0; the distance from (x,y) to this line is `|x-y+3|/sqrt(2) `

For y=-x+3 we rewrite as x+y-3=0; the distance from (x,y) to this line is `|x+y-3|/sqrt(2) `

Since the point is to be equidistant from the lines we set the expressions for the distance equal to each other:

`|x-y+3|/sqrt(2)=|x+y-3|/sqrt(2) `

Multiply each side by `sqrt(2) ` to get |x-y+3|=|x+y-3|

Now square each side -- this eliminates the absolute value bars, but might introduce extraneous solutions.

` x^2-xy+3x-xy+y^2-3y+3x-3y+9=x^2+xy-3x+xy+y^2-3y-3x-3y+9 ` Add like terms and bring to one side of the equation to get:

12x-4xy=0

4x(3-y)=0

So by the zero product property x=0 or y=3.

If you observe the given two lines

They will intersect at (0,3) on the y-axis.

So the two lines which are passing through the intersection point and perpendicular to each are the locus. They are x=0 and y=3.