In the XY-plane, are the points with their coordinates (a, b) and (c,

\({a^2} + {b^2}\,\,\mathop = \limits^? \,\,{c^2} + {d^2}\)


\(\left( 1 \right)\,\,a + b = 2\,\,\,\left\{ \matrix{\\
\,{\rm{Take}}\,\,\left( {a,b} \right) = \left( {1,1} \right)\,\,\,{\rm{and}}\,\,\,\left( {c,d} \right) = \left( {1,1} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{Yes}}} \right\rangle \hfill \cr \\
\,{\rm{Take}}\,\,\left( {a,b} \right) = \left( {1,1} \right)\,\,\,{\rm{and}}\,\,\,\left( {c,d} \right) = \left( {0,0} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{No}}} \right\rangle \hfill \cr} \right.\)


\(\left( 2 \right)\,\,\,\left\{ \matrix{\\
\,a + c = 1 \hfill \cr \\
\,b + d = 1 \hfill \cr} \right.\,\,\,;\,\,\,\,\left\{ \matrix{\\
\,{\rm{Take}}\,\,\left( {a,b} \right) = \left( {0.5,0.5} \right)\,\,\,{\rm{and}}\,\,\,\left( {c,d} \right) = \left( {0.5,0.5} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{Yes}}} \right\rangle \hfill \cr \\
\,\left( {{\mathop{\rm Re}\nolimits} } \right){\rm{Take}}\,\,\left( {a,b} \right) = \left( {1,1} \right)\,\,\,{\rm{and}}\,\,\,\left( {c,d} \right) = \left( {0,0} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{No}}} \right\rangle \hfill \cr} \right.\)



\(\left( {1 + 2} \right)\,\,\left\{ \matrix{\\
\,a + c = 1 \hfill \cr \\
\,b + d = 1 \hfill \cr} \right.\,\,\,\,\mathop \Rightarrow \limits^ + \,\,\,a + b + c + d = 2\,\,\,\,\mathop \Rightarrow \limits^{\left( 1 \right)} \,\,\,c + d = 0\)


\(\left\{ {\left. \matrix{\\
\,{a^2} + {b^2} = {\left( {2 - b} \right)^2} + {b^2} = 4 - 4b + 2{b^2} \hfill \cr \\
\,{c^2} + {d^2} = {\left( { - d} \right)^2} + {d^2} = 2{d^2} = 2{\left( {1 - b} \right)^2} = 2 - 4b + 2{b^2} \hfill \cr} \right\}\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{No}}} \right\rangle } \right.\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

Solution:

Using the distance formula, the distance between (a, b) and the origin is √[(a - 0)^2 + (b - 0)^2] = √(a^2 + b^2). Similarly, the distance between (c, d) and the origin is √[(c - 0)^2 + (d - 0)^2] = √(c^2 + d^2). We need to determine whether √(a^2 + b^2) =√(c^2 + d^2).

Statement One Alone:

a + b = 2

Since this doesn’t tell us anything about c and d, statement one alone is not sufficient.

Statement Two Alone:

c = 1 − a and d = 1 − b

We see that if a = 1/2 and b = 1/2, then c = 1 - a = 1/2 and d = 1 - b = 1/2. In this case, (a, b) = (c, d) = (1/2, 1/2) and thus, the two distances are equal. However, if a = 1 and b = 1, then c = d = 0. In this case, (a, b) = (1, 1), which has a positive distance to the origin; but (c, d) = (0, 0), which has a zero distance to the origin. Statement two alone is not sufficient.

Statements One and Two Together:

From statement two, we see that c^2 = (1 - a)^2 = a^2 - 2a + 1 and d^2 = (1 - b)^2 = b^2 - 2b + 1. Thus, the distance of (c, d) to the origin can be expressed in terms of a and b as follows:

c^2 + d^2

a^2 - 2a + 1 + b^2 - 2b + 1

a^2 + b^2 - 2(a + b) + 2

From statement one, we have a + b = 2. Let’s substitute 2 for a + b:

a^2 + b^2 - 2(2) + 2

a^2 + b^2 - 4 + 2

a^2 + b^2 - 2

Recall that the distance of (a, b) to the origin is a^2 + b^2, and the distance of (c, d) to the origin is c^2 + d^2 = a^2 + b^2 - 2. Since a^2 + b^2 - 2 is less than a^2 + b^2 no matter what values we pick for a and b, the distance between (c, d) and the origin is less than the distance between (a, b) and the origin. Statements one and two together are sufficient.

Answer: C

Why from statement B we cant use midpoint formula and prove that midpoint is not equal to origin?