Find n so that line is perpendicular to the line with the equation -2y+4=6x+8 through the points at (n,40) and (2, -8)
+23246 Line will be perpendicular if their slopes are negative reciprocal.
First, find the slope of -2y + 4 = 6x + 8
-2y = 6x + 4
y = -3x - 2
The slope of this line is -3 so the slope of any line parallel to it will be 1/3.
Now, find the equation of the line that passes through (2,-8) with a slope of 1/3.
Since we know a point and a slope, let's use the point-slope form: y - y1 = m(x - x1)
---> y - -8 = (1/3)(x - 2)
---> y + 8 = (1/3)(x - 2)
---> 3y + 24 = x - 2
---> -x + 3y = -26
---> x - 3y = 26 <--- This is the equation of the line perpendicular to y = -3x - 2 at the point (2, -8)
To find the value of n of the point (n, 40), replace y with 40:
x - 3(40) = 26
x - 120 = 26
x = 146 = n
+23246 Line will be perpendicular if their slopes are negative reciprocal.
First, find the slope of -2y + 4 = 6x + 8
-2y = 6x + 4
y = -3x - 2
The slope of this line is -3 so the slope of any line parallel to it will be 1/3.
Now, find the equation of the line that passes through (2,-8) with a slope of 1/3.
Since we know a point and a slope, let's use the point-slope form: y - y1 = m(x - x1)
---> y - -8 = (1/3)(x - 2)
---> y + 8 = (1/3)(x - 2)
---> 3y + 24 = x - 2
---> -x + 3y = -26
---> x - 3y = 26 <--- This is the equation of the line perpendicular to y = -3x - 2 at the point (2, -8)
To find the value of n of the point (n, 40), replace y with 40:
x - 3(40) = 26
x - 120 = 26
x = 146 = n
