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Question

What is the area (in sq. units) of the triangle formed by the graphs of the equations 2x + 5y - 12 = 0, x + y = 3 and y = 0?

This question was previously asked in
SSC CGL 2019 (Tier 2) GS Finance & Economics Previous Year Paper (17-Nov-2020)
The correct answer is

3 unit2

This explanation demonstrates how to calculate the area of a triangle formed by the intersection of three specific lines: 2x + 5y - 12 = 0, x + y = 3, and y = 0. We need to find the area in square units.

Finding Triangle Vertices

The vertices of the triangle are the points where each pair of the given lines intersect. We will find these three intersection points.

Vertex 1: Intersection of Lines x + y = 3 and y = 0

To find the intersection point, substitute the value of y from the second equation into the first equation:

Substitute \( y = 0 \) into \( x + y = 3 \):

\( x + 0 = 3 \)

This simplifies to \( x = 3 \).

So, the first vertex is (3, 0).

Vertex 2: Intersection of Lines 2x + 5y - 12 = 0 and y = 0

Similarly, substitute \( y = 0 \) into the equation \( 2x + 5y - 12 = 0 \):

\( 2x + 5(0) - 12 = 0 \)

This simplifies to \( 2x - 12 = 0 \).

Adding 12 to both sides gives \( 2x = 12 \).

Dividing by 2, we get \( x = 6 \).

So, the second vertex is (6, 0).

Vertex 3: Intersection of Lines 2x + 5y - 12 = 0 and x + y = 3

We have a system of two linear equations:

  • 2x + 5y - 12 = 0
  • x + y = 3

From the second equation, we can express x in terms of y: \( x = 3 - y \).

Now, substitute this expression for x into the first equation:

\( 2(3 - y) + 5y - 12 = 0 \)

Distribute the 2: \( 6 - 2y + 5y - 12 = 0 \)

Combine like terms: \( 3y - 6 = 0 \)

Add 6 to both sides: \( 3y = 6 \)

Divide by 3: \( y = 2 \).

Now, substitute the value of y back into the expression for x:

\( x = 3 - y \)

\( x = 3 - 2 \)

\( x = 1 \).

So, the third vertex is (1, 2).

Calculating Triangle Area

We have found the three vertices of the triangle: (3, 0), (6, 0), and (1, 2).

Notice that two vertices, (3, 0) and (6, 0), lie on the x-axis (the line \( y = 0 \)). We can use the segment connecting these points as the base of the triangle.

Base Calculation

The length of the base is the distance between the points (3, 0) and (6, 0).

\( \text{Base Length} = |x_2 - x_1| = |6 - 3| = 3 \) units.

Height Calculation

The height of the triangle is the perpendicular distance from the third vertex (1, 2) to the line containing the base (the x-axis). This is simply the absolute value of the y-coordinate of the third vertex.

\( \text{Height} = |y_3| = |2| = 2 \) units.

Area Calculation Using Base and Height

The formula for the area of a triangle is:

$$ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} $$

Substitute the calculated base and height:

$$ \text{Area} = \frac{1}{2} \times 3 \times 2 $$

$$ \text{Area} = \frac{1}{2} \times 6 $$

$$ \text{Area} = 3 $$

The area of the triangle is 3 square units.

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Similar Questions

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Important Questions from Co-ordinate Geometry

  1. The graphs of the linear equations 4x - 2y = 10 and 4x + ky = 2 intersect at a point (a, 4). The value of k is equal to:

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