# pre-cal

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Can someone please show me how to find an equation for the line with the given properties.

Parellel to the lines 3x-y= -3;
containing the point (0,0)

y=

• pre-cal -

If the new line is parallel to 3x-y = -3 it must differ only in the constant.
So the new equation is
3x - y = c
but (0,0) lies on it, so
3(0) - 0 = c
c = 0

new equation :
3x - y = 0
or
y = 3x

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