What is the equation of a line perpendicular to 8y-2x=-6 and also passing through the point (2,3).
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The product of the slope of perpendicular lines is -1. First find the slope of the given line.
8y - 2x = -6
=> 8y = 2x - 6
=> y = x/4 - 3/4
This is in the slope-intercept form y = mx + c where m is the slope.
The slope is 1/4. As a line perpendicular to this line has a slope which is the negative reciprocal, the slope of the required line is -4. The line also passes through (2,3)
The equation of the line is : (y - 3)/(x - 2) = -4
=> y - 3 = -4x + 8
=> 4x + y - 11 = 0
The equation of the required line is 4x + y - 11 = 0
We are asked to find the equation of a line perpendicular to
8y-2x = -6 which passes through the point (2,3).
We begin by finding the slope of the given line. We will convert the given equation to slope-intercept form.
=>8y - 2x = -6
=> 8y = 2x - 6
=> y = 2/8 x - 6/8
=> y = 1/4 x - 3/4
The slope of a line perpendicular to the above equation would be -4.
We substitute -4 as the perpendicular slope and the given point (2,3) into the slope intercept form to find "b."
=>y = mx + b
=> 3 = -4(2) + b
=> 3 = -8 + b
=> 11 = b
We now have the slope and the y intercept for the equation of the perpendicular line.
Substitute our values into the slope intercept form.
y = mx + b
y = -4x + 11.
The answer is y = -4x + 11 (slope-intercept form). The standard form of the answer is 4x + y = 11.
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