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In the xy-plane, the graphs of x=3
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10 Mar 2021, 05:09
10 Mar 2021, 05:09
1
00:00
Question Stats:
75% (01:09) correct
25% (01:36) wrong based on 8 sessions
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In the xy-plane, the graphs of x=3 and y = 6 intersect at point P, and the graphs of x=3 and y=-2 intersect at point Q. Origin O, point P, and point Q form a triangular region OPQ. what is the area OPQ?
(A)4
(B)8
(C)12
(D)16
(E)20
(A)4
(B)8
(C)12
(D)16
(E)20
Re: In the xy-plane, the graphs of x=3
[#permalink]
22 Jun 2021, 15:22
22 Jun 2021, 15:22
2
The question tells us that line x=3 intersects y=6 and y=-2 at points P and Q. Also these points with origin O are forming triangle.
Now to find the area of triangle we need to have the length of the base and height. "area of triangle = bh/2".
The base is PQ. Since P and Q have the same x-axis, the length of PQ is the difference of y values.
P=(3,6) and Q=(3,-2)
PQ length= 6-(-2)= 8
The height here is the distance from O "origin" to line x=3 or line PQ which is 3.
Area of triangle = (8*3)/2= 12.
Now to find the area of triangle we need to have the length of the base and height. "area of triangle = bh/2".
The base is PQ. Since P and Q have the same x-axis, the length of PQ is the difference of y values.
P=(3,6) and Q=(3,-2)
PQ length= 6-(-2)= 8
The height here is the distance from O "origin" to line x=3 or line PQ which is 3.
Area of triangle = (8*3)/2= 12.
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General Discussion
Re: In the xy-plane, the graphs of x=3
[#permalink]
10 Mar 2021, 05:49
10 Mar 2021, 05:49
Expert Reply
This is NOT a QCQ question 
Moved under PS forum
Moved under PS forum
