I need to know what the perpendicular equation is to x-3y=6 with the line through (-2,-9)
there is a perpendicular dis. between this line and pt. ie. 19|root of 10
or, since you have a point (-2,-9) and a slope (-3):
(y+9)/(x+2) = -3
To find the equation of the line perpendicular to the line given, you need to determine the slope of the given line first. The given line is in the form of "x - 3y = 6," which can be rewritten in slope-intercept form (y = mx + b), where m represents the slope.
Let's rearrange the given equation to make y the subject:
x - 3y = 6
-3y = -x + 6
y = (1/3)x - 2
In this equation, the coefficient of x (1/3) represents the slope of the line. The slope of the given line is therefore 1/3.
To find the slope of the line that is perpendicular to this line, you need to find the negative reciprocal of the given slope. The negative reciprocal of 1/3 is -3/1 or simply -3.
Now that you have the slope (-3) and a point through which the perpendicular line passes (-2, -9), you can use the point-slope form of a line to find the equation.
The point-slope form of a line is given by:
y - y1 = m(x - x1)
where (x1, y1) is the given point and m is the slope.
Plugging in the values we know:
y - (-9) = -3(x - (-2))
y + 9 = -3(x + 2)
y + 9 = -3x - 6
Finally, simplify the equation:
y = -3x - 6 - 9
y = -3x - 15
Therefore, the equation of the line perpendicular to x - 3y = 6 that passes through the point (-2, -9) is y = -3x - 15.