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# On the rectangular coordinate plane, points P, Q, and R form a right

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Director
Joined: 04 Dec 2015
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Location: India
Concentration: Technology, Strategy
Schools: ISB '19, IIMA , IIMB, XLRI
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On the rectangular coordinate plane, points P, Q, and R form a right [#permalink]

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11 Sep 2017, 05:20
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93% (01:55) correct 7% (01:43) wrong based on 68 sessions

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On the rectangular coordinate plane, points $$P$$, $$Q$$, and $$R$$ form a right triangle with two sides that are parallel to the $$x$$ and $$y$$ axes. If point $$P$$ is located at $$(-3, 2)$$, point $$R$$ is located at $$(3,-6)$$, and the $$90$$ degree angle is at point $$Q$$, what is the perimeter of triangle $$PQR$$?

(A) 10
(B) 12
(C) 14
(D) 17
(E) 24
[Reveal] Spoiler: OA
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On the rectangular coordinate plane, points P, Q, and R form a right [#permalink]

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11 Sep 2017, 06:04
If P(-3,2) and R(3,-6) are two points the distance between PR is $$\sqrt{6^2 + (-8)^2}$$ = 10

Slope(m) = $$\frac{y2 - y1}{x2 - x1} = \frac{-8}{6} = -\frac{4}{3}$$

The equation of the line is $$y - y1 = m(x - x1)$$
Substituting the values, the equation of the line PR is $$y + 6 = \frac{-4}{3}(x - 3) => 3y + 18 = -4x + 12 => 4x + 3y = -6$$

If y = 0,x = -1.5
Similarly if x = 0, y = -2
This equation has x and y intercepts and cannot be parallel to the axes.
This side PR is not parallel to the x and y axes and is the hypotenuse, with a measure of 10.

The other 2 sides of this right angled triangle are 6 and 8.
Therefore, the perimeter of this triangle is 6+8+10 = 24(Option E)
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Joined: 02 Jul 2017
Posts: 288
GMAT 1: 730 Q50 V38
Re: On the rectangular coordinate plane, points P, Q, and R form a right [#permalink]

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11 Sep 2017, 23:24
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KUDOS
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Here it will be easy if we draw the points on co-ordinate plane

points P, Q and R form a right triangle
with two sides that are parallel to the x and y axis <=Imp
P (−3,2) and R (3,−6) and 90 degree angle is at point Q
Find => perimeter of triangle PQR ?

So if we mark point P (-3,2) and R (3,-6) and we know sides are parallel to x and y axis.. just by diagram we know Q will be either (3,2) or (-3,-6) <- Refer attached figure

In both cases distance will remain constant. So to find length of triangle let just pick 1 point of Q (3,2)

So by distance formula
$$PQ = \sqrt{(-3-3)^2 + (2-2)^2} = \sqrt{(-6)^2} = 6$$
$$QR = \sqrt{(3-3)^2 + (-6-2)^2} = \sqrt{(-8)^2} = 8$$
PR we can find either by distance formula or using Pythagorean theorem as we know 2 sides.
$$PR = \sqrt{6^2 +8^2} = \sqrt{100} = 10$$

Perimeter PQR = 6+8+10 = 24

Attached image => points marked on co-ordinate plane.

Attachments

Right triangle coordinate plane.jpg [ 2.13 MiB | Viewed 725 times ]

Intern
Joined: 13 Sep 2017
Posts: 1
Re: On the rectangular coordinate plane, points P, Q, and R form a right [#permalink]

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13 Sep 2017, 19:46
1
KUDOS
Just by mapping the points on a coordinate scale and tracking the distance between the points will give us the length of the sides of the triangle.
In this case , the distance between -3 and 3 is 6 and similarly , the distance between -2 and 6 is 8.
Hence according to the Pythagoras theorem , the third side is 10
Simply by adding 6,8 and 10 , we get the required perimeter to be 24

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Joined: 22 May 2016
Posts: 1329
On the rectangular coordinate plane, points P, Q, and R form a right [#permalink]

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14 Sep 2017, 07:34
sashiim20 wrote:
On the rectangular coordinate plane, points $$P$$, $$Q$$, and $$R$$ form a right triangle with two sides that are parallel to the $$x$$ and $$y$$ axes. If point $$P$$ is located at $$(-3, 2)$$, point $$R$$ is located at $$(3,-6)$$, and the $$90$$ degree angle is at point $$Q$$, what is the perimeter of triangle $$PQR$$?

(A) 10
(B) 12
(C) 14
(D) 17
(E) 24

Attachment:

345tri.png [ 39.78 KiB | Viewed 544 times ]

I agree with KJags (kudos given!) :

Even without graph paper, this triangle is easy to draw.

Sketch, and list the coordinates of P and R.
Draw a straight line to the right from P and a straight line up from R.

The two lines meet at right-angled vertex Q.
Q's x-coordinate is the same as R
Q's y-coordinate is the same as P
Q is at (3,2)

Find the length of the sides PQ and QR
I don't like the distance formula; I find it cumbersome and prone to error.

Either count the length of each leg (PQ = 6 and QR = 8), OR

Find length of PQ by subtracting x-coordinates: 3 - (-3) = 6
Find length of QR by subtracting y-coordinates: 2 - (-6) = 8

This is a 3-4-5 right triangle, so 6: 8: 10 = hypotenuse = length of PR*
Perimeter is 6 + 8 + 10 = 24

*Alternatively use Pythagorean theorem:
$$6^2 + 8^2 = (PR)^2$$
$$100 = (PR)^2$$
$$PR = 10$$

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On the rectangular coordinate plane, points P, Q, and R form a right   [#permalink] 14 Sep 2017, 07:34
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