Finding the Value of m for a Specific Bounded Area

The problem asks us to find the value of \(m\), given that \(m < 0\), for which the area bounded by the lines \(y = x\), \(y = mx\), and \(x = 2\) is equal to 3.

Understanding the Bounded Region

We are given three lines:

• \(y = x\): A line passing through the origin with a slope of 1.
• \(y = mx\): A line passing through the origin with a slope of \(m\). Since \(m < 0\), this line has a negative slope.
• \(x = 2\): A vertical line at \(x = 2\).

The region bounded by these lines forms a triangle. Let's find the vertices of this triangle by determining the intersection points of these lines.

Finding Intersection Points

1. Intersection of \(y = x\) and \(y = mx\):

   Set the y-values equal: \(x = mx\)

   \(x - mx = 0\)

   \(x(1 - m) = 0\)

   Since \(m < 0\), \(m \neq 1\), so \(1 - m \neq 0\). This implies \(x = 0\). When \(x = 0\), \(y = 0\). The intersection point is \((0, 0)\).

2. Intersection of \(y = x\) and \(x = 2\):

   Substitute \(x = 2\) into \(y = x\): \(y = 2\). The intersection point is \((2, 2)\).

3. Intersection of \(y = mx\) and \(x = 2\):

   Substitute \(x = 2\) into \(y = mx\): \(y = m(2) = 2m\). The intersection point is \((2, 2m)\).

The vertices of the bounded region (the triangle) are \((0, 0)\), \((2, 2)\), and \((2, 2m)\).

Calculating the Area of the Triangle

The three vertices are \((0, 0)\), \((2, 2)\), and \((2, 2m)\). Notice that two vertices, \((2, 2)\) and \((2, 2m)\), lie on the vertical line \(x = 2\). We can use the segment connecting these two points as the base of the triangle.

• The length of the base is the absolute difference in the y-coordinates of \((2, 2)\) and \((2, 2m)\):

  Base length = \(|2 - 2m|\).

  Since \(m < 0\), \(2m\) is a negative number. For example, if \(m = -1\), \(2m = -2\). Then \(|2 - (-2)| = |4| = 4\). In general, \(2 - 2m\) will be positive when \(m < 0\). So, Base length = \(2 - 2m\).

• The height of the triangle with respect to this base is the perpendicular distance from the third vertex, \((0, 0)\), to the line \(x = 2\) (which contains the base). The distance from the origin \((0, 0)\) to the vertical line \(x = 2\) is the absolute value of the x-coordinate, which is \(|2| = 2\).

  Height = 2.

The area of a triangle is given by the formula: Area = \(\frac{1}{2} \times \text{base} \times \text{height}\).

Area = \(\frac{1}{2} \times (2 - 2m) \times 2\)

Area = \(2 - 2m\)

Solving for m

We are given that the area bounded by the lines is equal to 3.

So, we set the calculated area equal to 3:

\(2 - 2m = 3\)

Now, we solve this equation for \(m\):

\(-2m = 3 - 2\)

\(-2m = 1\)

\(m = -\frac{1}{2}\)

Checking the Condition

The problem states that \(m < 0\). Our calculated value is \(m = -\frac{1}{2}\). Since \(-\frac{1}{2} < 0\), this value satisfies the given condition.

Conclusion

The value of \(m\) for which the area bounded by the lines \(y = x\), \(y = mx\), and \(x = 2\) is equal to 3, with \(m < 0\), is \(m = -\frac{1}{2}\).

Revision Table: Area Bounded by Lines

Additional Information: Area Calculation Methods

Besides the geometric method used above (calculating the area of a triangle using base and height), the area bounded by curves can also be found using integral calculus.

• Using Integration:

  The area between two curves \(y = f(x)\) and \(y = g(x)\) from \(x = a\) to \(x = b\), where \(f(x) \ge g(x)\) over the interval \([a, b]\), is given by \(\int_{a}^{b} (f(x) - g(x)) dx\).

  In this problem, for \(x\) from 0 to 2, the upper curve is \(y = x\) and the lower curve is \(y = mx\) (since \(m < 0\), \(mx\) is negative or zero for \(x \ge 0\), while \(x\) is positive or zero for \(x \ge 0\). Thus \(x \ge mx\) for \(x \in [0, 2]\) when \(m < 0\)).

  Area = \(\int_{0}^{2} (x - mx) dx\)

  Area = \(\int_{0}^{2} x(1 - m) dx\)

  Area = \((1 - m) \int_{0}^{2} x dx\)

  Area = \((1 - m) \left[ \frac{x^2}{2} \right]_{0}^{2}\)

  Area = \((1 - m) \left( \frac{2^2}{2} - \frac{0^2}{2} \right)\)

  Area = \((1 - m) \left( \frac{4}{2} - 0 \right)\)

  Area = \((1 - m) \times 2\)

  Area = \(2 - 2m\)

  Setting this equal to 3: \(2 - 2m = 3\), which gives \(m = -\frac{1}{2}\).

• Using the Shoelace Formula (for polygon vertices):

  Given vertices \((x_1, y_1), (x_2, y_2), \dots, (x_n, y_n)\), the area is \(\frac{1}{2} |(x_1y_2 + x_2y_3 + \dots + x_ny_1) - (y_1x_2 + y_2x_3 + \dots + y_nx_1)|\).

  For vertices \((0, 0), (2, 2), (2, 2m)\), ordered counterclockwise or clockwise:

  Let's use \((0,0), (2,2), (2, 2m)\).

  Area = \(\frac{1}{2} |(0 \times 2 + 2 \times 2m + 2 \times 0) - (0 \times 2 + 2 \times 2 + 2m \times 0)|\)

  Area = \(\frac{1}{2} |(0 + 4m + 0) - (0 + 4 + 0)|\)

  Area = \(\frac{1}{2} |4m - 4|\)

  Area = \(\frac{1}{2} |4(m - 1)|\)

  Area = \(2 |m - 1|\)

  Since \(m < 0\), \(m - 1\) is negative. So \(|m - 1| = -(m - 1) = 1 - m\).

  Area = \(2 (1 - m) = 2 - 2m\).

  Setting this equal to 3: \(2 - 2m = 3\), which gives \(m = -\frac{1}{2}\).

  Both the geometric method, integration, and the shoelace formula yield the same result, \(m = -\frac{1}{2}\).