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08 Jul 2018, 21:40
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B
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Difficulty:
45%
(medium)
Question Stats:
59% (01:36) correct
41%
(01:29)
wrong
based on 317
sessions
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In the rectangular coordinate system above, the area of triangular region PQR is
A. \(\frac{ac}{2}\)
B. \(\frac{(c-a)(b-d)}{2}\)
C. \(\frac{(c+a)(b+d)}{2}\)
D. \(\frac{c(b-d)}{2}\)
E. \(\frac{c(b+d)}{2}\)
Attachment:
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08 Jul 2018, 21:59
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\(Area = \frac{1}{2} * base * height\)
Let the point where PR intersect x-axis is S. The co-ordinates of S will be (a,0)
the base of the triangle is PR
the height of the triangle is QS
Co-ordinates of P and Q are (a,b) and (a,d) respectively
base = |b| + |d| = b-d (since |b| = b as it is y co-ordinate in 2nd quadrant and |d|=-d as it is the y co-ordinate in 3rd quadrant)
height = |c| - |a| = c-a (since |c| = c as it is x co-ordinate in 1st quadrant and |a|=-a as it x co-ordinate is in 2nd quadrant)
Hence \(Area = \frac{(c-a)(b-d)}{2}\)
Hence option B
Let the point where PR intersect x-axis is S. The co-ordinates of S will be (a,0)
the base of the triangle is PR
the height of the triangle is QS
Co-ordinates of P and Q are (a,b) and (a,d) respectively
base = |b| + |d| = b-d (since |b| = b as it is y co-ordinate in 2nd quadrant and |d|=-d as it is the y co-ordinate in 3rd quadrant)
height = |c| - |a| = c-a (since |c| = c as it is x co-ordinate in 1st quadrant and |a|=-a as it x co-ordinate is in 2nd quadrant)
Hence \(Area = \frac{(c-a)(b-d)}{2}\)
Hence option B
General Discussion
