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a translation 5 units right and 3 units down, then a reflection across …

Question

a translation 5 units right and 3 units down, then a reflection across ( x = 0 ) is applied to the point ( (5, -1) ). where is the image in the coordinate plane?

a. quadrant i

b. quadrant ii

c. quadrant iii

d. quadrant iv

Explanation:

Step1: Apply Translation

The original point is \((5, -1)\). Translating 5 units right (add 5 to \(x\)-coordinate) and 3 units down (subtract 3 from \(y\)-coordinate):
New \(x = 5 + 5 = 10\), new \(y = -1 - 3 = -4\). So after translation, the point is \((10, -4)\).

Step2: Apply Reflection across \(x = 0\) (y-axis)

The rule for reflecting across \(x = 0\) is \((x, y) \to (-x, y)\).
Applying this to \((10, -4)\): \(x\)-coordinate becomes \(-10\), \(y\)-coordinate remains \(-4\). So the image is \((-10, -4)\).

Step3: Determine Quadrant

In the coordinate plane:

  • Quadrant I: \((+, +)\)
  • Quadrant II: \((-, +)\)
  • Quadrant III: \((-, -)\)
  • Quadrant IV: \((+, -)\)

The point \((-10, -4)\) has \((-, -)\), so it is in Quadrant III.

Answer:

C. Quadrant III