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

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

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07 Nov 2018, 02:08
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In the XY-plane, are the points with their coordinates (a, b) and (c, d) equidistant from the origin?

(1) a + b = 2
(2) c = 1 − a and d = 1 − b

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Re: In the XY-plane, are the points with their coordinates (a, b) and (c,  [#permalink]

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07 Nov 2018, 05:24
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Bunuel wrote:
In the XY-plane, are the points with their coordinates (a, b) and (c, d) equidistant from the origin?

(1) a + b = 2
(2) c = 1 − a and d = 1 − b

Algebraically:
If the points (a, b) and (c, d) are equidistant from the origin ( & say the distance is x), then $$a^2+b^2=x^2=c^2+d^2$$
1) No information about (c, d) ------ Not Suff
2) $$c^2+d^2=(1-a)^2 +(1-b)^2= 1- 2a + a^2 + 1 - 2b + b^2 = (a^2+b^2) + 2 - 2(a+b)$$
Clearly when in the aforementioned, $$a+b=2$$ the expression becomes $$c^2+d^2=x^2=a^2+b^2$$.... other wise not ....as two ans possible ----- Not Suff
1+2) Stmt 2 says : $$c^2+d^2 = (a^2+b^2) + 2 - 2(a+b)$$ & Stmt 01 says : $$a + b = 2$$
Thus , $$c^2+d^2 = (a^2+b^2) + 2 - 2(a+b) = (a^2+b^2) + 2 - 2*2 = (a^2+b^2) - 2 != (a^2+b^2)$$.... Thus Sufficient
Hence Ans C.

Geometrically : Trying to visualize Stmt 02 , not done yet.
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In the XY-plane, are the points with their coordinates (a, b) and (c,  [#permalink]

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02 Dec 2018, 10:38
Bunuel wrote:
In the XY-plane, are the points with their coordinates (a, b) and (c, d) equidistant from the origin?

(1) a + b = 2
(2) c = 1 − a and d = 1 − b

If the points (a, b) and (c, d) are equidistant from the origin then a^2+b^2=c^2+d^2
1) No information about (c, d) ------ Not sufficient.
2) c^2+d^2=(1−a)^2+(1−b)^2=1−2a+a^2+1−2b+b^2=(a^2+b^2)+2−2(a+b)
So to make it equal to a^2+b^2, we need 2-2(a+b)=0
So we get a+b=1 but no information is given. Not sufficient.

1+2) Stmt 1 says a+b=2 but for equidistant we want a+b=1....Not possible. Thus Sufficient

Hence Ans C.
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Re: In the XY-plane, are the points with their coordinates (a, b) and (c,  [#permalink]

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03 Dec 2018, 18:49
1
Bunuel wrote:
In the XY-plane, are the points with their coordinates (a, b) and (c, d) equidistant from the origin?

(1) a + b = 2
(2) c = 1 − a and d = 1 − b

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

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

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

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

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

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

Regards,
Fabio.
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Re: In the XY-plane, are the points with their coordinates (a, b) and (c,   [#permalink] 03 Dec 2018, 18:49
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