Write an equation for each line in point-slope form and then convert it to standard form
write an equation of the line parallel to x+2y=6 through (8,3)
write an equation of the line perpendicular to x+2y=6 through (8,3)
graph the three lines on the same coordinate plane

Step-by-step explanation:

I find it easier to work with the given standard-form equation. The parallel line will have the same x- and y-coefficients and a new constant. That constant can be found by substituting the given x- and y-values into the left-side expression:

  x + 2y = 8 + 2·3 = 14

The parallel line is x + 2y = 14.

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The perpendicular line will have the x- and y-coefficients swapped and one of them negated. (In standard form, the x-coefficient is positive, so in this case it is convenient to negate the y-coefficient.) Then the perpendicular line through (8, 3) is ...

  2x -y = 2·8 -3 = 13

The perpendicular line is 2x - y = 13.

check the picture below.

well, the angle at the vertex d is clearly not a right-angle, let's check for the one at the vertex f which does look like,   however if it's indeed, that means ef is perpendicular to df, in which case their slopes will be negative reciprocal of each other.

so the negative reciprocal of the slope of ef is 1, however the slope of df is not 1, is 1/2, so nope, they're not perpendicular lines.

x=6 and x=3

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