LINR 1 | Lesson 4 | Practice (Writing the Equation of Parallel and Perpendicular Lines Solutions)

1.  The slope of the line is \(3\).  The equation of the line parallel to \(y = 3x – 1\) and through the point \((1,-1)\) is \(y = 3x – 4\) in Slope-Intercept Form.


2.  The slope of the line is \(3\).  The slope of the line perpendicular to \(y=3x-2\) and through the point \((2, 4)\) is \(-\Large\frac{1}{3}\).  The equation is \(x + 3y = 14\) in Standard Form.


3.  The midpoint of the line segment is \((5, 1)\). The slope of the line segment is 0, and the slope of the perpendicular line is  \(-\Large\frac{1}{0}\) or undefined.  We know this to be a vertical line.  Therefore the equation is \(x = 5\).


4.  The midpoint of the segment is \((3, 5,)\).  The slope of the segment is \(1\). The slope of the line perpendicular to the line segment and through the point \((3,5)\) is \(-1\).  The equation of the perpendicular bisector in Standard Form is \(x+y = 8\).


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