# 2x+y-5=0;(0,4)write the standard form of the equation of the line that is parallel to the graph of the given equation and that passes through the point with the given coordinate.

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We are given the line `2x+y-5=0` and a point (0,4) and we are asked to find the equation of a line through the point that is parallel to the given line.

The slope of a line in standard form `Ax+By+C=0` is `m=-A/B` . Thus the slope of the given line is m=-2

The slope of parallel lines is the same, so we seek the line through the point (0,4) with slope -2. We can use the point slope form : Given a point `(x_1,y_1)` and a slope m, the equation is `y-y_1=m(x-x_1)`

So the equation is `y-4=-2(x-0)==>y-4=-2x`

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**In standard form the equation is `2x+y-4=0` **

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The graphs of teh two lines:

Here, equation of the given line: 2x+y-5=0

so slope of the line=-(coeficient of x/coefficient of y)

=-2/1

=-2

Slope of line parralel with this line is the same. Thus the slope of the required line(m)=-2

The line passes through (0,4),i.e, (x',y'). We have the formula to find the equation of a line when its slope is given and when a point on it is given as:

(y-y')=m(x-x')

or y-4=-2(x-0)

or y-4=-2x

or, y=-2x+4

This is the required equation which is in the form of y=mx+c.