ok im tring to find the slope intercept form of a perpindicular line so i can graph it heres the equation: (-3,1) y=-3x+7
It makes no sense to put a pair of coordinates (-3, 1) in front of a "y" term in an equation. Are you trying to find the equatin of a line that passes through (-3,1) and is perpendicular to
y = -3x+7 ?
That would not make sense either, becasuse y = -3x +7 does not pass through that point.
Is your new line perpendicular to y = -3x + 7 ?
If so, then the new slope must be (+1/3) and the new line is
y = (1/3)x + b
but (-3,1) is supposed to be on this new line, so
1 = (1/3)(-3) + b
1 = -1 + b
b = 2
new line : y = (1/3)x + 2
To find the equation of a perpendicular line, you need to determine the slope of the given line and then find the negative reciprocal of that slope.
In the given equation y = -3x + 7, the slope is -3, as it is the coefficient of x.
To find the negative reciprocal of -3, you take the reciprocal of -3, which is -1/3, and then multiply it by -1. Therefore, the negative reciprocal of -3 is 1/3.
Now that you have the slope of the perpendicular line, you can use the given point (-3, 1) and the slope intercept form (y = mx + b) to find the equation of the line.
Substituting the known values into the slope intercept form, you get:
1 = (1/3)(-3) + b
Simplifying, you have:
1 = -1 + b
To isolate b, add 1 to both sides:
b = 2
Therefore, the slope-intercept form of the perpendicular line is y = (1/3)x + 2.