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

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Director
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On the rectangular coordinate plane, points P, Q, and R form a right [#permalink]

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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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New post 11 Sep 2017, 07:04
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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Re: On the rectangular coordinate plane, points P, Q, and R form a right [#permalink]

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New post 12 Sep 2017, 00:24
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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.


Answer: E
Attachments

Right triangle coordinate plane.jpg
Right triangle coordinate plane.jpg [ 2.13 MiB | Viewed 552 times ]

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Re: On the rectangular coordinate plane, points P, Q, and R form a right [#permalink]

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New post 13 Sep 2017, 20:46
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The answer is (E) 24
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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On the rectangular coordinate plane, points P, Q, and R form a right [#permalink]

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New post 14 Sep 2017, 08:34
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
345tri.png [ 39.78 KiB | Viewed 374 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

Answer E

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

Add sides for perimeter

Kudos [?]: 337 [0], given: 591

On the rectangular coordinate plane, points P, Q, and R form a right   [#permalink] 14 Sep 2017, 08:34
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On the rectangular coordinate plane, points P, Q, and R form a right

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