Question

week 6 hw - parallel and perpendicular lines
geometry
name__ per__

1. write an equation of a horizontal line through (-3,2).
2. write an equation of a vertical line through the point (0,2).
3. write an equation of a line that is perpendicular to x = -1 and goes through (10,7).
4. write an equation of a line that is parallel to y = 7 and goes through (-1,-2).
5. line b has equation y = 9 - 10x.

a. circle the equation of the line that is parallel to line b.
y = -10x + 11 y = -\frac{1}{10}x + 4 y = -\frac{1}{9}x + 7 y = \frac{1}{10}x - \frac{1}{9}
b. circle the equation of the line that is perpendicular to line b.
y = -10x + 11 y = -\frac{1}{10}x + 4 y = -\frac{1}{9}x + 7 y = \frac{1}{10}x - \frac{1}{9}

1. are the following lines perpendicular? explain your reasoning.

Explanation:

Step1: Recall equation of horizontal line

The equation of a horizontal line is of the form $y = k$, where $k$ is the $y$-coordinate of any point on the line. For a horizontal line through $(-3,2)$, the $y$-coordinate is $2$. So the equation is $y=2$.

Step2: Recall equation of vertical line

The equation of a vertical line is of the form $x = h$, where $h$ is the $x$-coordinate of any point on the line. For a vertical line through $(0,2)$, the $x$-coordinate is $0$. So the equation is $x = 0$.

Step3: Analyze line perpendicular to $x=-1$

The line $x=-1$ is vertical. A line perpendicular to a vertical line is horizontal. For a horizontal line through $(10,7)$, the equation is $y = 7$ since the $y$-coordinate of the point is $7$.

Step4: Analyze line parallel to $y = 7$

The line $y = 7$ is horizontal. A line parallel to a horizontal line is also horizontal. For a horizontal line through $(-1,-2)$, the equation is $y=-2$ as the $y$-coordinate of the point is $-2$.

Step5: Find parallel line to $y=9 - 10x$

The slope - intercept form of a line is $y=mx + b$, where $m$ is the slope. For the line $y=9 - 10x$ (or $y=-10x + 9$), the slope $m=-10$. Parallel lines have the same slope. So the line parallel to it is $y=-10x + 11$.

Step6: Find perpendicular line to $y=9 - 10x$

If two lines with slopes $m_1$ and $m_2$ are perpendicular, then $m_1m_2=-1$. Given $m_1=-10$, then $m_2=\frac{1}{10}$. So the line perpendicular to $y = 9-10x$ is $y=\frac{1}{10}x-\frac{1}{9}$.

Step7: Check perpendicular lines on graph

Find the slopes of the two lines on the graph. If the product of their slopes is $-1$, they are perpendicular. Let's assume we find the slopes $m_1$ and $m_2$ of the two lines from the graph. If $m_1m_2=-1$, then they are perpendicular.

Answer:

1. $y = 2$
2. $x = 0$
3. $y = 7$
4. $y=-2$
5. a. $y=-10x + 11$

b. $y=\frac{1}{10}x-\frac{1}{9}$

1. Check the product of the slopes of the two lines on the graph. If the product is $-1$, they are perpendicular; otherwise, they are not.