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Apr 2808:00 AM PDT
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14 May 2019, 00:08
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
56% (03:15) correct
44%
(02:59)
wrong
based on 89
sessions
History
Date
Time
Result
Not Attempted Yet
Two lines with slopes \(\frac{1}{2}\) and 2 intersect at (2, 2). What is the area of the triangle enclosed by these two lines and the line x + y = 10 ?
(A) 4
(B) \(4\sqrt{2}\)
(C) 6
(D) 8
(E) \(6\sqrt{2}\)
(A) 4
(B) \(4\sqrt{2}\)
(C) 6
(D) 8
(E) \(6\sqrt{2}\)
14 May 2019, 02:40
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bit lengthy though
first - make equation of three lines
second -draw these three lines
third-find vertices of triangle by equating
last find area of triangle using coordinates (4,6),(2,2) and (6,4)
=1/2{X1(Y2-Y3)+X2(Y3-Y1)+X3(Y1-Y2)
i think ans is 6
any shortcut method would be appreciated
first - make equation of three lines
second -draw these three lines
third-find vertices of triangle by equating
last find area of triangle using coordinates (4,6),(2,2) and (6,4)
=1/2{X1(Y2-Y3)+X2(Y3-Y1)+X3(Y1-Y2)
i think ans is 6
any shortcut method would be appreciated
14 May 2019, 07:53
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Bunuel
Equation of Line with slope 1/2 and passing through (2,2)
y = (1/2)x+c
@(2,2), equation becomes 2 = (1/2)*2+c i.e. c = 1 and equation becomes
y = (1/2)*x+1
Equation of Line with slope 2 and passing through (2,2)
y = (2)x+c
@(2,2), equation becomes 2 = (2)*2+c i.e. c = -2 and equation becomes
y = (2)*x-2
Third line y = -x+10
Follow the colour coding for checking the graphs of respective lines
The points of intersection by solving the equations of lines we get
(2,2) (4,6) (6,4)
i.e. Sides of triangle are √20, √20 and √8=2√2
i.e. It's an isosceles triangle with base = √8 and height = √(20-2) = √18
Area = (1/2)*√18 * 2√2 = √36 = 6
Answer: Option C
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