What set of reflections would carry rectangle ABCD onto itself?
1 answer:
Answer:
y=x, x-axis, y=x, y-axis
Explanation:
Reflecting the figure across three axes just moves it from one quadrant to another. It does not map the figure to itself.
Reflecting across the line y=x moves it from quadrant II to IV or vice-versa. If it is in quadrant I or III, it stays there. So the sequence of reflections x-axis (moves from I to IV), y=x (moves from IV to II), x-axis (moves from II to III), y=x (stays in III) will not map the figure to itself.
However, the last selection will map the figure to itself. The initial (and final) figure location, and the intermediate reflections are shown in the attached. The figure starts and ends as blue, is reflected across y=x to green, across x-axis to orange, across y=x to red, and finally across y-axis to blue again.
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Answer:
Part A:
Hence proved that Vector= ai + bj is perpendicular to the line ax + by = c.
Part B:
Slope of vector =
Step-by-step explanation:
Condition for perpendicular is:
Part A:
Consider the vector v = ai + bj
x component of vector=a
y component of vector=b
Slope of vector=
Consider the line ax + by = c:
Rearranging the equation:
ax+by=c
by=c-ax
y=
According to general equation of line:
Where m is the slope
In our case the slope of above line is:
According to the condition of perpendicular:
Hence proved that Vector= ai + bj is perpendicular to the line ax + by = c.
Part B:
Slope of vector is also calculated above.
Since the slope of vector is negative reciprocal of the slope of the given line:
According to equation of line ax + by = c
y=
According to general equation of line:
Where m is the slope
Slope of given line=m=
negative reciprocal of the slope of the given line =
Slope of vector =