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The area of the triangle formed by the graphs of the equations x = 0, 2x + 3y = 6 and x + y = 3 is (SSC CGL 1st Sit. 2015)
- a)3 sq. unit
- b)1(1/2) sq. unit
- c)1 sq. unit
- d)4(1/2) sq. unit
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
The area of the triangle formed by the graphs of the equations x = 0, ...
To find the area of the triangle formed by the given equations, we need to find the vertices of the triangle and then use the formula for the area of a triangle.
Given equations:
1) x = 0
2) 2x - 3y = 6
3) x + y = 3
To find the vertices, we can solve the equations to find the points of intersection.
Solving equations 2) and 3) simultaneously:
2x - 3y = 6
x + y = 3
Multiplying the second equation by 2 to eliminate x:
2(x + y) = 2(3)
2x + 2y = 6
Now we have the system of equations:
2x - 3y = 6
2x + 2y = 6
Subtracting the equations to eliminate x:
(2x - 3y) - (2x + 2y) = 6 - 6
-5y = 0
y = 0
Substituting y = 0 into equation 2):
2x - 3(0) = 6
2x = 6
x = 3
The point of intersection of equations 2) and 3) is (3, 0).
Next, we need to find the point of intersection of equations 1) and 2).
Substituting x = 0 into equation 2):
2(0) - 3y = 6
-3y = 6
y = -2
The point of intersection of equations 1) and 2) is (0, -2).
Lastly, we need to find the point of intersection of equations 1) and 3).
Substituting x = 0 into equation 3):
0 + y = 3
y = 3
The point of intersection of equations 1) and 3) is (0, 3).
Now we have the three vertices of the triangle: (3, 0), (0, -2), and (0, 3).
To find the area of the triangle, we can use the formula for the area of a triangle given its vertices:
Area = 1/2 * |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|
Substituting the coordinates of the vertices into the formula:
Area = 1/2 * |3(3 - (-2)) + 0((-2) - 0) + 0(0 - 3)|
Area = 1/2 * |3(5) + 0 + 0|
Area = 1/2 * |15|
Area = 1/2 * 15
Area = 7.5
Therefore, the area of the triangle formed by the given equations is 7.5 square units, which is equivalent to option B).
Given equations:
1) x = 0
2) 2x - 3y = 6
3) x + y = 3
To find the vertices, we can solve the equations to find the points of intersection.
Solving equations 2) and 3) simultaneously:
2x - 3y = 6
x + y = 3
Multiplying the second equation by 2 to eliminate x:
2(x + y) = 2(3)
2x + 2y = 6
Now we have the system of equations:
2x - 3y = 6
2x + 2y = 6
Subtracting the equations to eliminate x:
(2x - 3y) - (2x + 2y) = 6 - 6
-5y = 0
y = 0
Substituting y = 0 into equation 2):
2x - 3(0) = 6
2x = 6
x = 3
The point of intersection of equations 2) and 3) is (3, 0).
Next, we need to find the point of intersection of equations 1) and 2).
Substituting x = 0 into equation 2):
2(0) - 3y = 6
-3y = 6
y = -2
The point of intersection of equations 1) and 2) is (0, -2).
Lastly, we need to find the point of intersection of equations 1) and 3).
Substituting x = 0 into equation 3):
0 + y = 3
y = 3
The point of intersection of equations 1) and 3) is (0, 3).
Now we have the three vertices of the triangle: (3, 0), (0, -2), and (0, 3).
To find the area of the triangle, we can use the formula for the area of a triangle given its vertices:
Area = 1/2 * |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|
Substituting the coordinates of the vertices into the formula:
Area = 1/2 * |3(3 - (-2)) + 0((-2) - 0) + 0(0 - 3)|
Area = 1/2 * |3(5) + 0 + 0|
Area = 1/2 * |15|
Area = 1/2 * 15
Area = 7.5
Therefore, the area of the triangle formed by the given equations is 7.5 square units, which is equivalent to option B).
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Community Answer
The area of the triangle formed by the graphs of the equations x = 0, ...
x = 0 ...(1)
2x + 3y = 6 ...(2)
x + y = 3 ...(3)
2x + 3y = 6 ...(ii)
x = 0, y = 2
x = 3, y = 0
From eqn. (iii) x+ y = 3
x = 0, y = 3
x = 3, y = 0
Area made by these three lines
= Area of triangle OBC – Area of OAC
2x + 3y = 6 ...(2)
x + y = 3 ...(3)
2x + 3y = 6 ...(ii)
x = 0, y = 2
x = 3, y = 0
From eqn. (iii) x+ y = 3
x = 0, y = 3
x = 3, y = 0
Area made by these three lines
= Area of triangle OBC – Area of OAC
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The area of the triangle formed by the graphs of the equations x = 0, 2x + 3y = 6 and x + y = 3 is (SSC CGL 1st Sit. 2015)a)3 sq. unitb)1(1/2) sq. unitc)1 sq. unitd)4(1/2)sq. unitCorrect answer is option 'B'. Can you explain this answer?
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The area of the triangle formed by the graphs of the equations x = 0, 2x + 3y = 6 and x + y = 3 is (SSC CGL 1st Sit. 2015)a)3 sq. unitb)1(1/2) sq. unitc)1 sq. unitd)4(1/2)sq. unitCorrect answer is option 'B'. Can you explain this answer? for SSC CGL 2024 is part of SSC CGL preparation. The Question and answers have been prepared according to the SSC CGL exam syllabus. Information about The area of the triangle formed by the graphs of the equations x = 0, 2x + 3y = 6 and x + y = 3 is (SSC CGL 1st Sit. 2015)a)3 sq. unitb)1(1/2) sq. unitc)1 sq. unitd)4(1/2)sq. unitCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for SSC CGL 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The area of the triangle formed by the graphs of the equations x = 0, 2x + 3y = 6 and x + y = 3 is (SSC CGL 1st Sit. 2015)a)3 sq. unitb)1(1/2) sq. unitc)1 sq. unitd)4(1/2)sq. unitCorrect answer is option 'B'. Can you explain this answer?.
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