Evgart wrote:
Bunuel wrote:
Official Solution:
For every point\((a, b)\) lying on line 1, point\((b, -a)\) lies on line 2. If the equation of line 1 is\(y = 2x + 1\) , what is the equation of line 2 ?
A. \(y = \frac{1}{2} + \frac{x}{2}\)
B. \(2y = 1 - x\)
C. \(\frac{x + y}{2} = -1\)
D. \(y = \frac{x}{2} - 1\)
E. \(x = 2y + 1\)
Find two points on line 2 and use their coordinates to build the line's equation. Points\((0, 1)\) and\((-\frac{1}{2}, 0)\) on
For every point\((a, b)\) lying on line 1, point\((b, -a)\) lies on line 2. If the equation of line 1 is\(y = 2x + 1\) , what is the equation of line 2 ?
A. \(y = \frac{1}{2} + \frac{x}{2}\)
B. \(2y = 1 - x\)
C. \(\frac{x + y}{2} = -1\)
D. \(y = \frac{x}{2} - 1\)
E. \(x = 2y + 1\)
Find two points on line 2 and use their coordinates to build the line's equation. Points\((0, 1)\) and\((-\frac{1}{2}, 0)\) on
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