Bunuel
In a rectangular coordinate system, are the points (x, y) and (w, z) equidistant from the origin?
(1) |x|+|y|=|w|+|z|
(2) x/y=w/z
Project DS Butler Data Sufficiency (DS3)
For DS butler Questions Click HereI'll change the letters in the question, because in coordinate geometry, x and y are usually reserved for variables (as in the equation of a line). So if we have the question:
Are (a, b) and (c, d) equidistant from the origin?
1. |a| + |b| = |c| + |d|
2. a/b = c/dNeither statement is sufficient alone; certainly the points can be equidistant from the origin, because they could be the same point. But for Statement 1 we could have the points (7, 0) and (3, 4), say, and by 3-4-5 triangles, (3, 4) is 5 units from the origin, while (7, 0) is 7 units from the origin. For Statement 2, one point could be (1, 1) and the other (1000, 1000), which are very different distances from (0, 0).
Using only Statement 2, conceptually it's easier to take reciprocals: b/a = d/c. Now b/a is equal to some number m. But if b/a = m, then b = ma, and the point (a, b) is on the line y = mx. Since b/a = d/c = m, we also have d = mc, and the point (c, d) is also on the line y = mx. So from Statement 2 alone, our two points are on the same line passing through the origin. And conceptually, if you start at the origin and travel along a line y = mx, the larger |x| gets, the further away from the origin you get. So if |a| + |b| = |c| + |d| here, then |a| = |c| must be true, and the points must be equidistant from (0, 0) if we use both Statements (they're either the same point, or one is the reflection through the origin of the other). Or you could do algebra: we know b = ma and d = mc, so substituting into Statement 1, we find
|a| + |b| = |c| + |d|
|a| + |ma| = |c| + |mc|
|a| + |m|*|a| = |c| + |m|*|c|
|a|(1 + |m|) = |c|(1 + |m|)
|a| = |c|
I've ignored some technicalities about zero, but if one of the x-coordinates is zero, the two Statements are clearly sufficient (and the question needs to rule out the possibility that the y-coordinates are zero since we divide by them in Statement 2).