# 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**