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11 Jul 2019, 23:23
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D
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Difficulty:
75%
(hard)
Question Stats:
57% (03:21) correct
43%
(02:50)
wrong
based on 76
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In the rectangular coordinate system, area of the region bounded by three lines x + y = 5, x - y = 7, y + 13x + 7 = 0 is:
A. 18
B. 22
C. 32
D. 42
E. 91
A. 18
B. 22
C. 32
D. 42
E. 91
12 Jul 2019, 01:34
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Bunuel
Let L1: x+y=5
L2: x-y=7
L3: y+13x+7=0
Intersection point of L1 & L2 is A (6,-1)
Intersection point of L3 & L2 is B (0,-7)
Intersection point of L1 & L3 is C (-1,6)
Distance between A & B = AB = base of triangle = \(\sqrt{(6-0)^2+(-1+7)^2} = 6\sqrt{2}\)
Height of the triangle ABC with base AB is = Perpendicular distance from C(-1,6) to Line L2 x-y-7=0 =\(|-1-6-7|/\sqrt{2} = 14/\sqrt{2} = 7\sqrt{2}\)
Area of triangle = base * height/2 = \(1/2 * 6\sqrt{2} * 7\sqrt{2} = \frac{6*7*2}{2} = 42\)
IMO D
12 Jul 2019, 02:41
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This is a very good question on Co-ordinate geometry, which tests you on the concept of Area of a triangle for which the co-ordinates of the vertices are given.
For a triangle whose vertices are (\(x_1,y_1\)), (\(x_2,y_2\)) and (\(x_3,y_3\)), as shown in the figure below,
12th July 2019 - Reply 4 - 1.JPG [ 33.64 KiB | Viewed 4732 times ]
Area of the triangle = ½ {\(x_1\)(\(y_2\)-\(y_3\)) + \(x_2\) (\(y_3\) – \(y_1\)) + \(x_3\) (\(y_1\) – \(y_2\))}. Observe that the expression on the right hand side is a cyclic expression and therefore easy to remember.
Since the question mentions the area bounded by 3 lines, in other words, it’s asking us to find out the area of a triangle.
Therefore, we need to find out the points where each pair of lines intersect.
The point of intersection of x + y = 5 and x – y = 7 is (6,-1).
The point of intersection of x + y = 5 and y + 13x + 7 = 0 is (-1,6).
The point of intersection of x – y = 7 and y + 13x + 7 = 0 is (0,-7).
These represent the three vertices of the triangle whose area we are trying to find, as shown in the figure below.
12th July 2019 - Reply 4 - 2.JPG [ 27.83 KiB | Viewed 4652 times ]
Let (\(x_1\),\(y_1\)) = (-1,6), (\(x_2\),\(y_2\)) = (0,-7) and (\(x_3\),\(y_3\)) = (6,-1). Substituting these values in the area formula,
Area of the triangle = ½ { -1(-7+1) + 0(-1-6) + 6(6+7)} = ½ {6 + 0 + 78} = 42 sq.units.
The correct answer option, IMO, is D.
Hope this helps!
For a triangle whose vertices are (\(x_1,y_1\)), (\(x_2,y_2\)) and (\(x_3,y_3\)), as shown in the figure below,
Attachment:
12th July 2019 - Reply 4 - 1.JPG [ 33.64 KiB | Viewed 4732 times ]
Area of the triangle = ½ {\(x_1\)(\(y_2\)-\(y_3\)) + \(x_2\) (\(y_3\) – \(y_1\)) + \(x_3\) (\(y_1\) – \(y_2\))}. Observe that the expression on the right hand side is a cyclic expression and therefore easy to remember.
Since the question mentions the area bounded by 3 lines, in other words, it’s asking us to find out the area of a triangle.
Therefore, we need to find out the points where each pair of lines intersect.
The point of intersection of x + y = 5 and x – y = 7 is (6,-1).
The point of intersection of x + y = 5 and y + 13x + 7 = 0 is (-1,6).
The point of intersection of x – y = 7 and y + 13x + 7 = 0 is (0,-7).
These represent the three vertices of the triangle whose area we are trying to find, as shown in the figure below.
Attachment:
12th July 2019 - Reply 4 - 2.JPG [ 27.83 KiB | Viewed 4652 times ]
Let (\(x_1\),\(y_1\)) = (-1,6), (\(x_2\),\(y_2\)) = (0,-7) and (\(x_3\),\(y_3\)) = (6,-1). Substituting these values in the area formula,
Area of the triangle = ½ { -1(-7+1) + 0(-1-6) + 6(6+7)} = ½ {6 + 0 + 78} = 42 sq.units.
The correct answer option, IMO, is D.
Hope this helps!
