In the standard \((\mathrm{x},\mathrm{y})\) coordinate plane, the graph of |3x| + |7y| = 21 is a parallelogram with 2 vertices...
GMAT Two Part Analysis : (TPA) Questions
In the standard \((\mathrm{x},\mathrm{y})\) coordinate plane, the graph of \(|\mathrm{3x}| + |\mathrm{7y}| = \mathrm{21}\) is a parallelogram with 2 vertices on the x-axis and 2 vertices on the y-axis.
Let \(\mathrm{H}\) and \(\mathrm{V}\) be the lengths, in coordinate units, of the horizontal and vertical diagonals, respectively, of this parallelogram. Select for \(\mathrm{H}\) and for \(\mathrm{V}\) values that are consistent with the given information.
Let's visualize this problem to make it crystal clear.
Phase 1: Understanding the Shape
The equation \(|3\mathrm{x}| + |7\mathrm{y}| = 21\) creates a special quadrilateral. Let's find its vertices by determining where it crosses the axes.
Finding x-axis intersections (where y = 0):
- \(|3\mathrm{x}| + |7(0)| = 21\)
- \(|3\mathrm{x}| = 21\)
- \(|\mathrm{x}| = 7\)
- So \(\mathrm{x} = 7\) or \(\mathrm{x} = -7\)
- Vertices on x-axis: \((7, 0)\) and \((-7, 0)\)
Finding y-axis intersections (where x = 0):
- \(|3(0)| + |7\mathrm{y}| = 21\)
- \(|7\mathrm{y}| = 21\)
- \(|\mathrm{y}| = 3\)
- So \(\mathrm{y} = 3\) or \(\mathrm{y} = -3\)
- Vertices on y-axis: \((0, 3)\) and \((0, -3)\)
Phase 2: Visualizing the Parallelogram
Let's draw this parallelogram:
(0,3)
|
|
(-7,0)---+---(7,0)
|
|
(0,-3)
The four vertices form a rhombus (special parallelogram) centered at the origin.
Phase 3: Finding the Diagonal Lengths
Horizontal diagonal H:
- Connects \((-7, 0)\) to \((7, 0)\)
- \(\mathrm{H} = 7 - (-7) = 14\) coordinate units
Vertical diagonal V:
- Connects \((0, -3)\) to \((0, 3)\)
- \(\mathrm{V} = 3 - (-3) = 6\) coordinate units
Phase 4: Solution
From our analysis:
- \(\mathrm{H} = 14\) (horizontal diagonal length)
- \(\mathrm{V} = 6\) (vertical diagonal length)
Answer:
- For H: Select 14
- For V: Select 6