What type of distribution is shown in the graph?
1 answer:
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We can start from the given line's coefficients and translate the line from the origin to the given point.
4(x -(-2)) -(y -3) = 0
4x +8 -y +3 = 0
The equation of the desired line is ...
4x -y = -11
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For standard form line ax+by=c, any parallel line will have only a different value of c. For c=0, the line goes through the origin (0, 0). To make it go through point (h, k) we can write it as
a(x-h) +b(y-k) = 0
which is completely equivalent to
ax +by = ah +bk
4(x -(-2)) -(y -3) = 0
4x +8 -y +3 = 0
The equation of the desired line is ...
4x -y = -11
_____
For standard form line ax+by=c, any parallel line will have only a different value of c. For c=0, the line goes through the origin (0, 0). To make it go through point (h, k) we can write it as
a(x-h) +b(y-k) = 0
which is completely equivalent to
ax +by = ah +bk