Question

The equation of a line passing through the point (2, -3) and parallel to 2x-y+3=0​

Answer 1

First we find the slope of the line 2x−3y+8=0 by placing it into slope intercept form:

2x−3y+8=0

⇒−3y=−2x−8

⇒3y=2x+8

⇒y=32x+38

 Therefore, the slope of the line is m=32.

Now since the equation of the line with slope m passing through a point (x1,y1) is

y−y1=m(x−x1)

Here the point is (2,3) and slope is m=32, therefore, the equation of the line is:

y−3=32(x−2)⇒3(y−3)=2(x−2)⇒3y−9=2x−4⇒2x−3y=−9+4⇒2x−3y=−5

Hence, the equation of the line is 2x−3y=−5.

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