The question asks for the equation of the line perpendicular to the given line that contains the point $P$.

Let's analyze and solve the problems one by one from the image.

Problem 19:

• Point: $P(6, 6)$
• Equation of the given line: $y = \frac{2}{3}x$

Solution for Problem 19:

1. Slope of the given line: The slope $m_1$ of the given line $y = \frac{2}{3}x$ is $\frac{2}{3}$.
2. Slope of the perpendicular line: The slope $m_2$ of a line perpendicular to another is the negative reciprocal of the original slope. So, $m_2 = -\frac{3}{2}$.
3. Equation of the line: Using the point-slope form of the line $y - y_1 = m(x - x_1)$, with point $P(6, 6)$ and slope $m_2 = -\frac{3}{2}$: $y - 6 = -\frac{3}{2}(x - 6)$
4. Simplify: $y - 6 = -\frac{3}{2}x + 9$ $y = -\frac{3}{2}x + 15$ So, the equation of the perpendicular line is: $y = -\frac{3}{2}x + 15$

Problem 20:

• Point: $P(4, 0)$
• Equation of the given line: $y = \frac{1}{2}x - 5$

Solution for Problem 20:

1. Slope of the given line: The slope $m_1$ of the line $y = \frac{1}{2}x - 5$ is $\frac{1}{2}$.
2. Slope of the perpendicular line: The slope of the perpendicular line will be the negative reciprocal, $m_2 = -2$.
3. Equation of the line: Using the point $P(4, 0)$ and slope $m_2 = -2$: $y - 0 = -2(x - 4)$
4. Simplify: $y = -2x + 8$ So, the equation of the perpendicular line is: $y = -2x + 8$

Problem 21:

• Point: $P(4, 4)$
• Equation of the given line: $y = -2x - 8$

Solution for Problem 21:

1. Slope of the given line: The slope $m_1$ of the line $y = -2x - 8$ is $-2$.
2. Slope of the perpendicular line: The slope of the perpendicular line will be the negative reciprocal, $m_2 = \frac{1}{2}$.
3. Equation of the line: Using the point $P(4, 4)$ and slope $m_2 = \frac{1}{2}$: $y - 4 = \frac{1}{2}(x - 4)$
4. Simplify: $y - 4 = \frac{1}{2}x - 2$ $y = \frac{1}{2}x + 2$ So, the equation of the perpendicular line is: $y = \frac{1}{2}x + 2$

Would you like more details on any part of the solution? Here are some questions for further practice:

1. How do you find the slope of a perpendicular line given the original slope?
2. How can you verify that two lines are perpendicular using their slopes?
3. What is the point-slope form of a line, and when is it useful?
4. How can you convert point-slope form into slope-intercept form?
5. How would you approach this problem if the point didn't lie directly on the given line?

Tip: Always check that the new line is indeed perpendicular by verifying that the product of the slopes of two perpendicular lines equals -1.