# 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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