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25 Jun 2020, 23:49
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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:
65%
(hard)
Question Stats:
63% (02:59) correct
38%
(02:45)
wrong
based on 40
sessions
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2.jpg [ 14.82 KiB | Viewed 2233 times ]
On the xy-plane above, if the equation of \(l_{1}\) is \(y=\frac{1}{2}x\) and if point B is defined by the xy-coordinate pair (5,0), what is the area of OAB ?
(A) 4
(B) 3\(\sqrt{2}\)
(C) 2\(\sqrt{5}\)
(D) 5
(E) 7
26 Jun 2020, 00:36
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Explanation:
Slope of line Ol1 is 1/2. Since angle at A is 90 degree, So slope of AB must be -2.
Draw a line from Pt.A towards OB , It will divide it by OB into 4 to 1. Height of Triangle OAB is 2.
Area = 1/2 x 2 x 5
=5
IMO-D
Slope of line Ol1 is 1/2. Since angle at A is 90 degree, So slope of AB must be -2.
Draw a line from Pt.A towards OB , It will divide it by OB into 4 to 1. Height of Triangle OAB is 2.
Area = 1/2 x 2 x 5
=5
IMO-D
28 Jun 2020, 06:00
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Let us follow a more concept oriented approach for this sum. It may seem long, but it really isn't.
Area of the Triangle = ½ * Base * Height
The first aim is to get the coordinates of A, and then the lengths of OA and AB
If the base is AB, then the height is the perpendicular OA
We are given that O is (0,0), B is (5,0) and OA is the line y = 1/2x ----- (1) .
The slope of OA = m1 = 1/2
Since OA is perpendicular to AB, therefore the product of the slopes = -1 i.e m1 x m2 = -1
Therefore m2 = -2
In line AB, we use the slope point form, y - y1 = m(x - x1) to find the equation of the line
Slope = -2, and the line passes through (5,0). Therefore y – 0 = -2(x – 5)
Or y = -2x + 10 ----- (2)
Since OA and AB intersect, solving equations (1) and (2), we get the co-ordinates of A
y = ½ x = - 2x + 10
2.5x = 10
x = 4
y = 1/2x = 4/2 = 2
Therefore the co ordinates of A are (4,2)
Using the distance formula between 2 points A (x1,y1) and B (x2,y2) = Sqrt(x1 – x2)^2 + (y1 – y2)^2:
OA = Sqrt[(4 – 0)^2 + (2 – 0)^2] = Sqrt(20)
AB = Sqrt[(5 – 4)^2 + (0 – 2)^2] = Sqrt(5)
Area of Triangle AOB = ½ * sqrt(20) * sqrt(5) = ½ * sqrt(100) = ½ * 10 = 5
Area of the Triangle = ½ * Base * Height
The first aim is to get the coordinates of A, and then the lengths of OA and AB
If the base is AB, then the height is the perpendicular OA
We are given that O is (0,0), B is (5,0) and OA is the line y = 1/2x ----- (1) .
The slope of OA = m1 = 1/2
Since OA is perpendicular to AB, therefore the product of the slopes = -1 i.e m1 x m2 = -1
Therefore m2 = -2
In line AB, we use the slope point form, y - y1 = m(x - x1) to find the equation of the line
Slope = -2, and the line passes through (5,0). Therefore y – 0 = -2(x – 5)
Or y = -2x + 10 ----- (2)
Since OA and AB intersect, solving equations (1) and (2), we get the co-ordinates of A
y = ½ x = - 2x + 10
2.5x = 10
x = 4
y = 1/2x = 4/2 = 2
Therefore the co ordinates of A are (4,2)
Using the distance formula between 2 points A (x1,y1) and B (x2,y2) = Sqrt(x1 – x2)^2 + (y1 – y2)^2:
OA = Sqrt[(4 – 0)^2 + (2 – 0)^2] = Sqrt(20)
AB = Sqrt[(5 – 4)^2 + (0 – 2)^2] = Sqrt(5)
Area of Triangle AOB = ½ * sqrt(20) * sqrt(5) = ½ * sqrt(100) = ½ * 10 = 5
