how do I find the locus of points in a plane equidistant from the lines y = x and y = -x ?

Answer 1

If what you mean by locus of points as the collection of points which satisfies y=x and y=-x then the the set is all rational numbers, i.e., y = any positive number and y = any negative number. If you were to graph this solution , it would like a cartesian coordinate plane with every point possible as the solution. Now if we throw in the term equidistant, the solution becomes limited to the line y=0, i.e., all rational numbers that are the same distance from one parallel line y = +x and the other parallel line y = -x. Graphing this solution looks like a line on the cartesian coordinate right on the x axis going through the point (0,0). Hope this helps.

Answer 2

The points equidistant from 2 intersecting lines (which form an angle) form the angle bisectors. Since y = x makes a 45° angle (up to the right) with the x axis, and y = -x makes a 45° angle (down right) with the x axis, they are perpendicular, and the x and y axes are the angle bisectors.

Answer 3

y = mx - a million has one primary aspect (0,-a million) that doesn't position self assurance in m. hit upon the intersection of both primary strains, and that is the distinctive aspect that the y=mx-a million line ought to wish to contain. The slope between those 2 elements is m. the primary strains intersect at the same time as both strains have the similar y fee for a given x, or: 4x + 4 = -4x + 12 8x = 8 x = a million both equations grant y=8 at the same time as x=a million, so (a million,8) is the intersection. From (0,-a million) to (a million,8) the slope is: m = (8 - (-a million)) / (a million - 0) = 9

Answer 4

y=x would be parallel to y= -x. for the locus points to be equidistant from both, they have to be in between the lines. since x and -x are equidistant from the y-axis, the locus points would be on the y-axis

Answer 5

these are the x and y axes