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On the xy-plane, if the point (4,t) is equidistant from the points (1,

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Bunuel
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On the xy-plane, if the point (4,t) is equidistant from the points (1, [#permalink] New post   17 Oct 2020, 13:16
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On the xy-plane, if the point (4,t) is equidistant from the points (1,1) and (5,3), then t =

(A) 3/2
(B) 1
(C) 0
(D) -1/4
(E) -4
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C
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On the xy-plane, if the point (4,t) is equidistant from the points (1, [#permalink] New post   17 Oct 2020, 21:48
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Bunuel wrote:
On the xy-plane, if the point (4,t) is equidistant from the points (1,1) and (5,3), then t =

(A) 3/2
(B) 1
(C) 0
(D) -1/4
(E) -4


The distant from point (4,t) to (1,1) is \(\sqrt{(4-1)^2+(t-1)^2}\)=\(\sqrt{9+t^2-2t+1}\)=\(\sqrt{t^2-2t+10}\)
The distant from point (4,t) to (5,3) is \(\sqrt{(4-5)^2+(t-3)^2}\)=\(\sqrt{1+t^2-6t+9}\)=\(\sqrt{t^2-6t+10}\)
\(\sqrt{t^2-2t+10}\)=\(\sqrt{t^2-6t+10}\) ---> t^2-2t+10=t^2-6t+10 ---> 4t=0 ---->t=0 C)
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Re: On the xy-plane, if the point (4,t) is equidistant from the points (1, [#permalink] New post   19 Oct 2020, 14:20
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Bunuel wrote:
On the xy-plane, if the point (4,t) is equidistant from the points (1,1) and (5,3), then t =

(A) 3/2
(B) 1
(C) 0
(D) -1/4
(E) -4


It's easier to plug in the answers instead of doing the problem algebraically. You can use the distance formula to calculate the distances, but that is not really necessary.

A) 1.5
→ (1, 1) to (4, 1.5) → right 3 and up 0.5
→ (5, 3) to (4, 1.5) → left 1 and down 1.5
Not equal

B) 1
→ (1, 1) to (4, 1) → right 3 and up 0
→ (5, 3) to (4, 1) → left 1 and down 2
Not equal

C) 0
→ (1, 1) to (4, 0) → right 3 and down 1
→ (5, 3) to (4, 0) → left 1 and down 3
The distances must be equal because the changes are both 3 and 1.
Equal
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Re: On the xy-plane, if the point (4,t) is equidistant from the points (1, [#permalink] New post   19 Oct 2020, 17:20
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Bunuel wrote:
On the xy-plane, if the point (4,t) is equidistant from the points (1,1) and (5,3), then t =

(A) 3/2
(B) 1
(C) 0
(D) -1/4
(E) -4


The midpoint of (1, 1) and (5, 3) is (3, 2). All points that are equidistant from those two points form a line containing (3, 2), this line is perpendicular to the line formed by (1, 1) and (5, 3) which has a slope of \(\frac{{3-1}}{{5-1}} = \frac{1}{2}\). If the perpendicular line has a slope of m then \(m * \frac{1}{2}= -1\).

Then m = -2 with (3, 2) on the line. To find (4, t) we move right by one unit, then move down by 1*2 = 2 units since the slope is -2. Then the point is (4, 0).

Ans: C
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Re: On the xy-plane, if the point (4,t) is equidistant from the points (1, [#permalink]
19 Oct 2020, 17:20
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