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amanvermagmat
If points A, B, C lie in the xy plane, is the area of triangle ABC greater than 6 square units?

(1) Equation of line AB is y=4 and equation of line BC is x=3.

(2) All three points A, B, C lie in first quadrant only.

Ans C: as axes do not come in any quadrant. As per both (1) and (2).... 0<x<=3 and 0<y<=4..... so area will be always less than 6 if the triangle lies in 1st quadrant
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amanvermagmat
If points A, B, C lie in the xy plane, is the area of triangle ABC greater than 6 square units?

(1) Equation of line AB is y=4 and equation of line BC is x=3.

(2) All three points A, B, C lie in first quadrant only.

Ans C: as axes do not come in any quadrant. As per both (1) and (2).... 0<x<=3 and 0<y<=4..... so area will be always less than 6 if the triangle lies in 1st quadrant

Hi durgamadhab ,
Your point is correct. However, the following are the points to ponder:-

1. In order to form a triangle , we have to consider the point of intersection of the lines AB & BC one of the vertices of triangle ABC. So, 'B' is the fixed vertices with co-ordinates B(3,4).

2. The points A & C have to lie necessarily on the line AB & BC respectively. (otherwise st1 is not validated)

3. ABC is a right angled triangle in the 1st quadrant.(As per st2)

4. With a fixed B; 0<AB<3 & 0<BC< infinity (subject to \(AC^2=AB^2+BC^2\)) (extending line BC vertically upward, C tends to have infinite no of positions))

5. Area of right angled triangle=\(\frac{1}{2}*AB*BC\) can have more than one value of square units.

Hope this is sufficient to say our answer E.
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durgamadhab
amanvermagmat
If points A, B, C lie in the xy plane, is the area of triangle ABC greater than 6 square units?

(1) Equation of line AB is y=4 and equation of line BC is x=3.

(2) All three points A, B, C lie in first quadrant only.

Ans C: as axes do not come in any quadrant. As per both (1) and (2).... 0<x<=3 and 0<y<=4..... so area will be always less than 6 if the triangle lies in 1st quadrant

Hi durgamadhab ,
Your point is correct. However, the following are the points to ponder:-

1. In order to form a triangle , we have to consider the point of intersection of the lines AB & BC one of the vertices of triangle ABC. So, 'B' is the fixed vertices with co-ordinates B(3,4).

2. The points A & C have to lie necessarily on the line AB & BC respectively. (otherwise st1 is not validated)

3. ABC is a right angled triangle in the 1st quadrant.(As per st2)

4. With a fixed B; 0<AB<3 & 0<BC< infinity (subject to \(AC^2=AB^2+BC^2\)) (extending line BC vertically upward, C tends to have infinite no of positions))

5. Area of right angled triangle=\(\frac{1}{2}*AB*BC\) can have more than one value of square units.

Hope this is sufficient to say our answer E.


When u take y=4, x=3 as two sides...and say the triangle will stay in first quadrant....the limits are fixed. maximum area of the triangle is certain.
Answer is C

another way....the area of the rectangle formed by lines y=4 and x=3 is.....3x4 = 12.
So the triangle will have a maximum area of 1/2 x area of rectangle...which is 1/2x12=6
But the points cannot be on the axes..... so the max area is 1/2 x 2.999999 x 3.999999= something less than 6
Ans C
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When u take y=4, x=3 as two sides...and say the triangle will stay in first quadrant....the limits are fixed. maximum area of the triangle is certain.
Answer is C[/quote]

y is NOT RESTRICTED ; it's range:- (0, INFINITY), which lie in 1st quadrant. So, we can't say what is the maximum area.

Given, y=4 is a line.
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When u take y=4, x=3 as two sides...and say the triangle will stay in first quadrant....the limits are fixed. maximum area of the triangle is certain.
Answer is C

y is NOT RESTRICTED ; it's range:- (0, INFINITY), which lie in 1st quadrant. So, we can't say what is the maximum area.

Given, y=4 is a line.[/quote]


Answer is E..... but thats because there is no limit to the x-coordinate on the line y=4 and to y-coordinate of x=3. The triangle can be on any of the 4 sides of point of intersection (3,4)
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Can we get an expert's comments on this? I got the OA because with my reading of the question, I identified point b as (3,4), saw that ABC was a right triangle, and assumed that the triangle could be formed on any of the four sides of point (3,4) formed by lines y=4 and x=3. Looking at others' comments, I wonder if there is any reason why the triangle has to be inside of (3,4) as opposed to extending potentially infinitely outward?
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This is a very good trap question. Caught me here.
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