How do you find a standard form equation for the line with (1,6) ; parallel to the line x+2y=6?

To find the standard form equation for the line parallel to ( x + 2y = 6 ) passing through the point ( (1,6) ), we first need to find the slope of the given line. We rearrange the equation into slope-intercept form, ( y = mx + b ), where ( m ) is the slope:

[ x + 2y = 6 ] [ 2y = -x + 6 ] [ y = -\frac{1}{2}x + 3 ]

The slope of the given line is ( -\frac{1}{2} ). Since the line we're looking for is parallel, it will have the same slope.

Now, using the point-slope form of a line, ( y - y_1 = m(x - x_1) ), where ( (x_1, y_1) ) is the given point and ( m ) is the slope:

[ y - 6 = -\frac{1}{2}(x - 1) ]

Expand and rearrange to get the equation in standard form:

[ 2(y - 6) = - (x - 1) ] [ 2y - 12 = -x + 1 ] [ x + 2y = 13 ]

So, the standard form equation for the line parallel to ( x + 2y = 6 ) passing through the point ( (1,6) ) is ( x + 2y = 13 ).