Question

geometry in construction
name_
id: 2
u2-d1,d2 quiz
period_
find the slope of each line.
1)
find the slope of the line through each pair of points.

1. $(-8,-19),(-15,9)$

find the slope of a line parallel to each given line.

1. $y=-2x - 2$

find the slope of a line perpendicular to each given line.

1. $y=\frac{1}{2}x + 1$

find the missing side of each triangle. round your answers to the nearest tenth if necessary.
5)
6)

Explanation:

Step1: Identify two points on the line (for problem 1)

From the graph, two points can be identified as \((-2,-4)\) and \((0,0)\)

Step2: Use the slope formula \(m=\frac{y_2 - y_1}{x_2 - x_1}\)

Substitute \((x_1,y_1)=(-2,-4)\) and \((x_2,y_2)=(0,0)\) into the formula: \(m=\frac{0 - (-4)}{0 - (-2)}=\frac{4}{2}=2\)

Step3: For problem 2, use the slope formula \(m=\frac{y_2 - y_1}{x_2 - x_1}\)

Here \((x_1,y_1)=(-8,-19)\) and \((x_2,y_2)=(-15,9)\)
\(m=\frac{9-(-19)}{-15 - (-8)}=\frac{9 + 19}{-15 + 8}=\frac{28}{-7}=-4\)

Step4: For problem 3, use the property of parallel lines

The equation of a line is \(y = mx + b\), where \(m\) is the slope. For \(y=-2x - 2\), the slope \(m=-2\). Parallel lines have the same slope, so the slope of the parallel line is \(-2\)

Step5: For problem 4, use the property of perpendicular lines

If a line has slope \(m_1\), a perpendicular line has slope \(m_2=-\frac{1}{m_1}\). For \(y=\frac{1}{2}x + 1\), \(m_1=\frac{1}{2}\), then \(m_2=-2\)

Step6: For problem 5, use the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\) (here \(c\) is the hypotenuse)

\(x=\sqrt{6.7^{2}+9^{2}}=\sqrt{44.89 + 81}=\sqrt{125.89}\approx11.2\)

Step7: For problem 6, use the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\) (here \(c = 15.8\), \(a = 6.9\))

\(x=\sqrt{15.8^{2}-6.9^{2}}=\sqrt{(15.8 + 6.9)(15.8 - 6.9)}=\sqrt{22.7\times8.9}=\sqrt{202.03}\approx14.2\)

Answer:

1. \(2\)
2. \(-4\)
3. \(-2\)
4. \(-2\)
5. \(11.2\) km
6. \(14.2\) yd