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sashiim20
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Schools: ISB '19 IIMB IIMA XLRI HEC Sept19 intake Rotman '21 NUS '21 NTU '20 Trinity MBA '20 Bocconi '21 Smurfit "21
Posts: 620
11 Sep 2017, 06:20
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E
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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
(A) 10
(B) 12
(C) 14
(D) 17
(E) 24
11 Sep 2017, 07:04
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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)
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)
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
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 [ 2.13 MiB | Viewed 3675 times ]
