arjtryarjtry wrote:
On the coordinate plane is point (0, 0) closer to point (u, v) than to point (u, v + 1) ?
(1) v + u^2 = -1
(2) v < 0
M22-11
On the coordinate plane is point (0, 0) closer to point (u, v) than to point (u, v + 1) ?The formula to calculate the distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\).
So basically the question asks whether the distance between the points \((0, 0)\) and \((u, v)\) is less than the distance between the points \((0, 0)\) and \((u, v + 1)\): is \(\sqrt{(u-0)^2+(v-0)^2}<\sqrt{(u-0)^2+(v+1-0)^2}\)? --> is \(\sqrt{u^2+v^2}<\sqrt{u^2+(v+1)^2}\)? --> is \(u^2+v^2<u^2+v^2+2v+1\)? --> is \(v>-\frac{1}{2}\)?
(1) \(v + u^2 = -1\) --> \(v=-1-u^2\leq{-1}\) --> so the answer to the question is NO. Sufficient.
(2) \(v<0\). Not sufficient.
Answer: A.
i could not understand the last step.
v=-1-u^2\leq{-1} --> so the answer to the question is NO.
I did not get how did you infer 'v' lesser than -1 from the last equation??