Determine the equation of the line that passes through the point (-2,4) and it is perpendicular to -2x+4y-1=0.
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The required line is perpendicular to -2x + 4y - 1 = 0
-2x + 4y - 1 = 0
=> 4y = 2x + 1
=> y = x/2 + 1/4
This is of the form y = mx + c where the slope is m and the y-intercept is c. Therefore the slope is 1/2.
The required line is perpendicular to -2x + 4y - 1 = 0, so its slope is the negative reciprocal of 1/2 or -2.
The equation of a line through (-2 , 4) and with a slope -2 is
(y - 4) / (x + 2) = -2
=> y - 4 = -2x - 4
=> 2x + y = 0
The required line is 2x + y = 0
We'll write the equation of the line into the slope intercept form:
y=mx+n, where m is the slope of the line and n is the y intercept.
We need to put the equations in this form to determine their slopes. We'll use the property of slopes of 2 perpendicular lines: the product of the values of the slopes of 2 perpendicular lines is -1.
Let's suppose that the 2 slopes are m1 and m2.
m1*m2=-1
We'll determine m1 from the given equation of the line, that is perpendicular to the one with the unknown equation.
The equation is -2x+4y-1=0.
We'll isolate 4y to the left side. For this reason, we'll subtract -2x - 1 both sides:
4y = 2x + 1
We'll divide by 4:
y = x/2 + 1/4
The slope m1 = 1/2.
(1/2)*m2=-1
m2=-2
We also know that the line passes through the point (-2,4), so the equation of a line that passes throuh a given point and it has a known slope is:
(y-y1)=m(x-x1)
(y-4)=(-2)*(x+2)
We'll remove the brackets and we'll move all terms to one side:
y - 4 + 2x + 4 = 0
We'll eliminate like terms and we'll get the equation of the requested line:
y + 2x = 0
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