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# In the figure above, what is the distance from point P to point Q?

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Retired Moderator
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In the figure above, what is the distance from point P to point Q? [#permalink]

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12 May 2015, 13:09
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71% (00:50) correct 29% (00:46) wrong based on 161 sessions

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In the figure above, what is the distance from point P to point Q?

A. √2
B. 1.52−1
C. √(1.52−1)
D. (√2)/2
E. 0.5

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Re: In the figure above, what is the distance from point P to point Q? [#permalink]

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12 May 2015, 21:53
Hi reto,

Any time that you have a diagonal line segment on a graph, you can "build" a right triangle around it (using the diagonal as the hypotenuse of the triangle). In this way, you can use the Pythagorean Theorem to calculate the length of the line segment.

Here, we have the co-ordinates (1, 1) and (1.5, 1.5). Drawing a right triangle here would give us a base of .5 and a height of .5

Using the Pythagorean Theorem, we have...

A^2 + B^2 = C^2
(.5)^2 + (.5)^2 = C^2
.25 + .25 = C^2
.5 = C^2

From here, you have to do a little work to "clean up" the answer

1/2 = C^2

Root(1/2) = C

Root(1/2) = (Root1)/(Root2) = 1/(Root2)

As a general rule, radicals in the denominator of a fraction are a "no-no", so you have to do the necessary math to 'get rid' of any radicals in the denominator.....

With 1/(Root2), we can multiply both the numerator and denominator by (Root2). This gives us...

(Root2)/[(Root2)^2] = (Root2)/2

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Re: In the figure above, what is the distance from point P to point Q? [#permalink]

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12 Sep 2017, 00:57
1
Given Q (1.5, 1.5) and P(1,1)
Find distance between P and Q.

Using distance formula as A(x1,y1) and B (x2,y2)
Distance =$$\sqrt{(x2-x1)^2 +(y2-y1)^2}$$

$$So distance between Q and P = \sqrt{(1.5-1)^2 +(1.5-1)^2} = \sqrt{(0.5)^2+(0.5)^2} = \sqrt{2 * (0.5)^2} = 0.5* \sqrt{2}$$

$$distance = 0.5* \sqrt{2} = \frac{5}{10} * \sqrt{2} = \frac{1}{2} * \sqrt{2}$$

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In the figure above, what is the distance from point P to point Q? [#permalink]

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13 Sep 2017, 02:25
1
reto wrote:
Attachment:
T6198.png
In the figure above, what is the distance from point P to point Q?

A. √2
B. 1.52−1
C. √(1.52−1)
D. (√2)/2
E. 0.5

The distance between the points is the hypotenuse of a 45-45-90 right triangle. The x- and y-coordinates both have a difference of 0.5.

Or: make a right triangle, with right leg going straight down from point (1.5, 1.5), and base (other leg) going straight across from (1,1). The lines will meet at (1.5, 1).

It's an isosceles right triangle. Each side is (1.5 - 1) = $$\frac{1}{2}$$.

As a 45-45-90 triangle, it has sides in ratio $$x: x: x\sqrt{2}$$.

The hypotenuse, the distance between the two points, therefore is $$(\frac{1}{2}) *(\sqrt{2})$$, or $$\frac{\sqrt{2}}{2}$$

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In the figure above, what is the distance from point P to point Q?   [#permalink] 13 Sep 2017, 02:25
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