Free Math Worksheets

- Identify the pairs of parallel lines first. - Find the vertices by intersecting one line from each parallel pair. - What is the vertical distance between the two horizontal lines? - Use the parallelogram area formula once you know a base and its corresponding height.

1. Find the vertices from pairwise intersections. From \(2 = x + 1\), \(f\) and \(h\) meet at \((1, 2)\). From \(2 = x - 4\), \(f\) and \(k\) meet at \((6, 2)\). From \(-3 = x - 4\), \(g\) and \(k\) meet at \((1, -3)\). From \(-3 = x + 1\), \(g\) and \(h\) meet at \((-4, -3)\). 2. The horizontal side from \((1, 2)\) to \((6, 2)\) has length \(6 - 1 = 5\). 3. The perpendicular height is the vertical distance between \(y = 2\) and \(y = -3\), which is \(2 - (-3) = 5\). 4. Therefore, the area is \(A = 5 \cdot 5 = 25\) square units.

\(25\) square units

- What does equal slope tell you about two distinct lines? - What x-coordinate does every point on the y-axis have? - Find where \(f\) and \(h\) cross the x-axis. - Use the distance between those x-intercepts as the base of the triangle.

1. The lines \(f\) and \(g\) have the same slope \(\frac{1}{2}\) but different y-intercepts, so they are parallel and do not intersect. 2. The y-intercept of \(f\) is \((0, 1)\). For \(h\) to pass through this point, \(h(0) = b = 1\). 3. The x-intercept of \(f\) satisfies \(\frac{1}{2}x + 1 = 0\), so \(x = -2\). The x-intercept of \(h(x) = -x + 1\) is \(x = 1\). 4. The base on the x-axis has length \(1 - (-2) = 3\). The height is the y-coordinate of the common point \((0, 1)\), so the height is \(1\). 5. The area is \(A = \frac{1}{2} \cdot 3 \cdot 1 = 1.5\) square units.

The graphs of \(f\) and \(g\) are parallel because they have the same slope and different y-intercepts. \(b = 1\). The triangle has area \(1.5\) square units.

- One of the given points already lies on a coordinate axis. Which one? - How can you find a line equation from two points? - What shape do the two intercepts and the origin form? - How do you find the area of a right triangle?

1. The slope through \(A\) and \(B\) is \(\frac{0 - 3}{4 - (-2)} = -\frac{1}{2}\). 2. Since \(B(4, 0)\) lies on the x-axis, the x-intercept is \((4, 0)\). 3. Write the line as \(y = -0.5x + b\). Using \((4, 0)\), \(0 = -0.5 \cdot 4 + b\), so \(b = 2\). The y-intercept is \((0, 2)\). 4. The axes and the line form a right triangle with legs of lengths \(4\) and \(2\). Its area is \(\frac{1}{2} \cdot 4 \cdot 2 = 4\) square units.

The intercepts are \((4, 0)\) and \((0, 2)\). The area is \(4\) square units.

- Find where the two lines have the same y-value. - The two y-intercepts form one side of the triangle. - What is the horizontal distance from the intersection point to the y-axis? - How does reflection affect area?

1. Set the line equations equal: \(x + 1 = -0.5x + 4\). Then \(1.5x = 3\), so \(x = 2\). Substitution gives \(y = 3\), so \(S = (2, 3)\). 2. The y-intercepts are \((0, 1)\) for line \(a\) and \((0, 4)\) for line \(b\). The vertical base of the triangle therefore has length \(4 - 1 = 3\). 3. The perpendicular distance from \(S(2, 3)\) to the y-axis is \(2\), so the triangle's height is \(2\). 4. The triangle's area is \(A = \frac{1}{2} \cdot 3 \cdot 2 = 3\) square units. 5. Reflection preserves area. The kite consists of the original triangle and its reflected copy, so its area is \(2 \cdot 3 = 6\) square units.

a) \(S = (2, 3)\) b) \(3\) square units c) \(6\) square units