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Point A (– 4, 2) and Point B (2, 4) lie in the xy-coordinate
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01 Dec 2018, 00:09
01 Dec 2018, 00:09
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Question Stats:
57% (02:02) correct
42% (02:28) wrong based on 80 sessions
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Point A (– 4, 2) and Point B (2, 4) lie in the xy-coordinate plane. If point C lies in the first quadrant and contains the coordinates (p, q), where p < 2 and q < 4, which of the following could be the area of triangle ABC?
Indicate all such numbers.
A. 1.1
B. 3.9
C. 11.9
Indicate all such numbers.
A. 1.1
B. 3.9
C. 11.9
Re: Point A (– 4, 2) and Point B (2, 4) lie in the xy-coordinate
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07 Jan 2019, 13:30
07 Jan 2019, 13:30
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Consider the following diagram the point C can be anywhere in the shaded region.
Lets take two extreme cases when the point C is 2,0 and when point C is 2,4. Now I know it is said that point c is p,q and p is less than 0 but we can select p = 1.99999999999 which is almost equal to 2 and q as 3.99999.
In the first case the area of the triangle is = base \(\frac{1}{2}\times\)\(\times\) height = \(\frac{1}{2} \times BC \times\) perpendicular distance of point A from BC = \(\frac{1}{2} \times 4 \times 6 =12\).
In the second case where point C is 2,4 length of BC is 0. So the area of the triangle is almost 0.
Clearly the area of triangle can go higher than 11.2 and as low as 0.
Hence all options are correct!
Lets take two extreme cases when the point C is 2,0 and when point C is 2,4. Now I know it is said that point c is p,q and p is less than 0 but we can select p = 1.99999999999 which is almost equal to 2 and q as 3.99999.
In the first case the area of the triangle is = base \(\frac{1}{2}\times\)\(\times\) height = \(\frac{1}{2} \times BC \times\) perpendicular distance of point A from BC = \(\frac{1}{2} \times 4 \times 6 =12\).
In the second case where point C is 2,4 length of BC is 0. So the area of the triangle is almost 0.
Clearly the area of triangle can go higher than 11.2 and as low as 0.
Hence all options are correct!
General Discussion
Re: Point A (– 4, 2) and Point B (2, 4) lie in the xy-coordinate
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01 Dec 2018, 09:42
01 Dec 2018, 09:42
Not so hard it just takes some time to understand and calculate the maximum area.
