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07 Jan 2022, 05:15
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
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Question Stats:
58% (02:57) correct
42%
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wrong
based on 90
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Given line L (illustrated in graph), and a parallel line that runs through point (-1,5), what is the perimeter of a rectangle whose sides run along the two lines and has a length of 9?
A. 28
B. \(9*\sqrt{15}\)
C. \(18+2*\sqrt{20}\)
D. 36
E. 45
Attachment:
1.png [ 47.72 KiB | Viewed 3283 times ]
10 Jan 2022, 04:02
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To solve, one first needs to find the distance between the two parallel lines.
First create a perpendicular line which goes between the coordinate \((-1 , 5)\) and Line L. As this line will be perpendicular to L it will have a negative reciprocal gradient of L's gradient. As L has a gradient of \(\frac{3}{4}\) then the perpendicular line will have a gradient of \(\frac{-4}{3}\).
So the perpendicular line will be \(y = \frac{-4}{3}x + c\). Plugging in \((-1 , 5)\) for x and y will let us solve for c.
\(5 = \frac{-4}{3} * (-1) + c\)
\(5 - \frac{4}{3} = + c\)
\(c = \frac{11}{3}\)
The full formula for the perpendicular line is: \(y = \frac{-4}{3}x + \frac{11}{3}\)
Next, make line L and the perpendicular line equal one another to find the coordinate at which they intersect, which can then be used in the distance formula together with coordinate \((-1 , 5)\)
\(\frac{3}{4}x - \frac{1}{2} = \frac{-4}{3}x + \frac{11}{3}\)
\(\frac{3}{4}x + \frac{4}{3}x = \frac{11}{3} + \frac{1}{2}\)
\(\frac{25}{12}x = \frac{25}{6}\)
\(x = 2\)
Plug that back into \(y = \frac{3}{4}x - \frac{1}{2}\) to get:
\(y = \frac{3}{4}(2) - \frac{1}{2}\)
\(y = \frac{6}{4} - \frac{1}{2}\)
\(y = \frac{4}{4}\)
\(y = 1\)
So the corresponding coordinate to \((-1 , 5)\) is \((2 , 1)\)
Distance formula: \(\sqrt{(X_2 - X_1)^2 + (Y_2 - Y_1)^2}\)
Plugging in the two coordinates to get their distance:
\(\sqrt{(-1 - 2)^2 + (5 - 1)^2}\)
= \(\sqrt{9 + 16}\)
= \(\sqrt{25}\)
= \(5\)
This gives us the length of one side, so the perimeter of the rectangle will be:
\((9*2) + (5*2)\)
=\(18 + 10\)
= \(28\)
Answer A
First create a perpendicular line which goes between the coordinate \((-1 , 5)\) and Line L. As this line will be perpendicular to L it will have a negative reciprocal gradient of L's gradient. As L has a gradient of \(\frac{3}{4}\) then the perpendicular line will have a gradient of \(\frac{-4}{3}\).
So the perpendicular line will be \(y = \frac{-4}{3}x + c\). Plugging in \((-1 , 5)\) for x and y will let us solve for c.
\(5 = \frac{-4}{3} * (-1) + c\)
\(5 - \frac{4}{3} = + c\)
\(c = \frac{11}{3}\)
The full formula for the perpendicular line is: \(y = \frac{-4}{3}x + \frac{11}{3}\)
Next, make line L and the perpendicular line equal one another to find the coordinate at which they intersect, which can then be used in the distance formula together with coordinate \((-1 , 5)\)
\(\frac{3}{4}x - \frac{1}{2} = \frac{-4}{3}x + \frac{11}{3}\)
\(\frac{3}{4}x + \frac{4}{3}x = \frac{11}{3} + \frac{1}{2}\)
\(\frac{25}{12}x = \frac{25}{6}\)
\(x = 2\)
Plug that back into \(y = \frac{3}{4}x - \frac{1}{2}\) to get:
\(y = \frac{3}{4}(2) - \frac{1}{2}\)
\(y = \frac{6}{4} - \frac{1}{2}\)
\(y = \frac{4}{4}\)
\(y = 1\)
So the corresponding coordinate to \((-1 , 5)\) is \((2 , 1)\)
Distance formula: \(\sqrt{(X_2 - X_1)^2 + (Y_2 - Y_1)^2}\)
Plugging in the two coordinates to get their distance:
\(\sqrt{(-1 - 2)^2 + (5 - 1)^2}\)
= \(\sqrt{9 + 16}\)
= \(\sqrt{25}\)
= \(5\)
This gives us the length of one side, so the perimeter of the rectangle will be:
\((9*2) + (5*2)\)
=\(18 + 10\)
= \(28\)
Answer A
10 Jan 2022, 04:25
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Did this. It took me more than 2 mins. Is it the case with everyone or just me? Is there a faster way to solve this.
