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26 Nov 2022, 08:15
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
62% (01:28) correct
38%
(01:43)
wrong
based on 13
sessions
History
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Time
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Not Attempted Yet
In the figure, the line l and the x-axis are tangent to the circle at points P and S, respectively, and the line segment QS passed through center R of the circle. What is the slope of l?
A. 1/2
B. √3/3
C. √3/2
D. 1
E. √3
Attachment:
2022-11-25_18-35-39.png [ 16.2 KiB | Viewed 1316 times ]
26 Nov 2022, 08:31
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In the figure, the line l and the x-axis are tangent to the circle at points P and S, respectively, and the line segment QS passed through center R of the circle. What is the slope of l?
A. 1/2
B. √3/3
C. √3/2
D. 1
E. √3
Solution is not difficult in this one but the ratios in the answer choices make it difficult.
Angle P = 90
Angle S = 90
Then,
Angle Q = 45
Thus, slope(QS/OS) makes and angle of 45. But since trigonometry is not tested we can can use equation of line y = mx + c where m is slope(QS/OS in this case)
Therefore, from pythagoras theorem.
QS = OS
QS/OS = 1
Answer D.
26 Nov 2022, 12:28
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Let the radius of the circle be r
In \(\triangle RPQ\)
Draw a line parallel to x axis passing through R. The line cuts the line l at point X.
In \(\triangle PRQ\)
Slope of the line = Rise / Run = \(\frac{\sqrt{2}x }{\sqrt{2}x}\) = 1
Option D
In \(\triangle RPQ\)
- \(\angle RPQ = 90^{\circ}\)
- \(\angle PRQ = 45^{\circ}\)
- \(\angle PQR = 45^{\circ}\)
- QR = \(\sqrt{2}r\) (using 45 - 45 - 90)
Draw a line parallel to x axis passing through R. The line cuts the line l at point X.
In \(\triangle PRQ\)
- \(\angle RPX = 90^{\circ}\)
- \(\angle PRX = 45^{\circ}\)
- \(\angle PXR = 45^{\circ}\)
- XR = \(\sqrt{2}r\) (using 45 - 45 - 90)
Slope of the line = Rise / Run = \(\frac{\sqrt{2}x }{\sqrt{2}x}\) = 1
Option D
Attachments
Screenshot 2022-11-27 005103.png [ 157.93 KiB | Viewed 1110 times ]
