Study Guide: Distance, Midpoint, and Graphing Equations in College Algebra

Q15. For the points $P(-5, -6)$ and $Q(7, -1)$, find (a) the distance $d(P, Q)$ and (b) the coordinates of the midpoint $M$ of line segment $PQ$.

Background

Topic: Distance and Midpoint Formulas in the Coordinate Plane

This question tests your ability to use the distance formula to find the length between two points and the midpoint formula to find the center point of a segment connecting two points.

Key formulas:

• Distance formula: $d(P, Q) = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

• Midpoint formula: $M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$

Step-by-Step Guidance

1. Identify the coordinates: $P(-5, -6)$ and $Q(7, -1)$, so $x_1 = -5$, $y_1 = -6$, $x_2 = 7$, $y_2 = -1$.

2. Set up the distance formula: $d(P, Q) = \sqrt{(7 - (-5))^2 + (-1 - (-6))^2}$

3. Calculate the differences: $7 - (-5)$ and $-1 - (-6)$, then square each result.

4. Set up the midpoint formula: $M = \left( \frac{-5 + 7}{2}, \frac{-6 + (-1)}{2} \right)$

Try solving on your own before revealing the answer!

Final Answer:

(a) $d(P, Q) = \sqrt{(7 - (-5))^2 + (-1 - (-6))^2} = \sqrt{(12)^2 + (5)^2} = \sqrt{144 + 25} = \sqrt{169} = 13$

(b) $M = \left( \frac{-5 + 7}{2}, \frac{-6 + (-1)}{2} \right) = \left( \frac{2}{2}, \frac{-7}{2} \right) = (1, -3.5)$

The distance between the points is 13 units, and the midpoint is at (1, -3.5).

Q16. For the points $P(-4, 3)$ and $Q(2, -5)$, find (a) the distance $d(P, Q)$ and (b) the coordinates of the midpoint $M$ of line segment $PQ$.

Background

Topic: Distance and Midpoint Formulas in the Coordinate Plane

This question tests your ability to use the distance formula and midpoint formula for two points in the plane.

Key formulas:

• Distance formula: $d(P, Q) = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

• Midpoint formula: $M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$

Step-by-Step Guidance

1. Identify the coordinates: $P(-4, 3)$ and $Q(2, -5)$, so $x_1 = -4$, $y_1 = 3$, $x_2 = 2$, $y_2 = -5$.

2. Set up the distance formula: $d(P, Q) = \sqrt{(2 - (-4))^2 + (-5 - 3)^2}$

3. Calculate the differences: $2 - (-4)$ and $-5 - 3$, then square each result.

4. Set up the midpoint formula: $M = \left( \frac{-4 + 2}{2}, \frac{3 + (-5)}{2} \right)$

Try solving on your own before revealing the answer!

Final Answer:

(a) $d(P, Q) = \sqrt{(2 - (-4))^2 + (-5 - 3)^2} = \sqrt{(6)^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$

(b) $M = \left( \frac{-4 + 2}{2}, \frac{3 + (-5)}{2} \right) = \left( \frac{-2}{2}, \frac{-2}{2} \right) = (-1, -1)$

The distance between the points is 10 units, and the midpoint is at (-1, -1).

Q17. For the points $P(8, 2)$ and $Q(3, 5)$, find (a) the distance $d(P, Q)$ and (b) the coordinates of the midpoint $M$ of line segment $PQ$.

Background

Topic: Distance and Midpoint Formulas in the Coordinate Plane

This question tests your ability to use the distance formula and midpoint formula for two points in the plane.

Key formulas:

• Distance formula: $d(P, Q) = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

• Midpoint formula: $M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$

Step-by-Step Guidance

1. Identify the coordinates: $P(8, 2)$ and $Q(3, 5)$, so $x_1 = 8$, $y_1 = 2$, $x_2 = 3$, $y_2 = 5$.

2. Set up the distance formula: $d(P, Q) = \sqrt{(3 - 8)^2 + (5 - 2)^2}$

3. Calculate the differences: $3 - 8$ and $5 - 2$, then square each result.

4. Set up the midpoint formula: $M = \left( \frac{8 + 3}{2}, \frac{2 + 5}{2} \right)$

Try solving on your own before revealing the answer!

Final Answer:

(a) $d(P, Q) = \sqrt{(3 - 8)^2 + (5 - 2)^2} = \sqrt{(-5)^2 + (3)^2} = \sqrt{25 + 9} = \sqrt{34} \approx 5.83$

(b) $M = \left( \frac{8 + 3}{2}, \frac{2 + 5}{2} \right) = \left( \frac{11}{2}, \frac{7}{2} \right) = (5.5, 3.5)$

The distance between the points is approximately 5.83 units, and the midpoint is at (5.5, 3.5).

Q18. For the points $P(-8, 4)$ and $Q(3, -5)$, find (a) the distance $d(P, Q)$ and (b) the coordinates of the midpoint $M$ of line segment $PQ$.

Background

Topic: Distance and Midpoint Formulas in the Coordinate Plane

This question tests your ability to use the distance formula and midpoint formula for two points in the plane.

Key formulas:

• Distance formula: $d(P, Q) = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

• Midpoint formula: $M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$

Step-by-Step Guidance

1. Identify the coordinates: $P(-8, 4)$ and $Q(3, -5)$, so $x_1 = -8$, $y_1 = 4$, $x_2 = 3$, $y_2 = -5$.

2. Set up the distance formula: $d(P, Q) = \sqrt{(3 - (-8))^2 + (-5 - 4)^2}$

3. Calculate the differences: $3 - (-8)$ and $-5 - 4$, then square each result.

4. Set up the midpoint formula: $M = \left( \frac{-8 + 3}{2}, \frac{4 + (-5)}{2} \right)$

Try solving on your own before revealing the answer!

Final Answer:

(a) $d(P, Q) = \sqrt{(3 - (-8))^2 + (-5 - 4)^2} = \sqrt{(11)^2 + (-9)^2} = \sqrt{121 + 81} = \sqrt{202} \approx 14.21$

(b) $M = \left( \frac{-8 + 3}{2}, \frac{4 + (-5)}{2} \right) = \left( \frac{-5}{2}, \frac{-1}{2} \right) = (-2.5, -0.5)$

The distance between the points is approximately 14.21 units, and the midpoint is at (-2.5, -0.5).

Q47. For the equation $y = \frac{1}{2}x - 2$, (a) give a table with at least three ordered pairs that are solutions, and (b) graph the equation.

Background

Topic: Linear Equations and Graphing

This question tests your ability to generate ordered pairs from a linear equation and plot them on a coordinate plane.

Key terms and formulas:

• Linear equation: $y = mx + b$ where $m$ is the slope and $b$ is the y-intercept.

• Ordered pair: $(x, y)$ values that satisfy the equation.

Step-by-Step Guidance

1. Choose three values for $x$ (e.g., $x = -2, 0, 2$).

2. Plug each $x$ value into the equation to solve for $y$.

3. Record each $(x, y)$ pair in a table.

4. Plot the points on a coordinate plane and draw a straight line through them.

Try solving on your own before revealing the answer!

Final Answer:

Sample table:

• $x = -2$, $y = \frac{1}{2}(-2) - 2 = -1 - 2 = -3$ → $(-2, -3)$

• $x = 0$, $y = \frac{1}{2}(0) - 2 = 0 - 2 = -2$ → $(0, -2)$

• $x = 2$, $y = \frac{1}{2}(2) - 2 = 1 - 2 = -1$ → $(2, -1)$

Graph: Plot these points and draw a straight line through them.

Q48. For the equation $y = -\frac{1}{2}x + 2$, (a) give a table with at least three ordered pairs that are solutions, and (b) graph the equation.

Background

Topic: Linear Equations and Graphing

This question tests your ability to generate ordered pairs from a linear equation and plot them on a coordinate plane.

Key terms and formulas:

• Linear equation: $y = mx + b$ where $m$ is the slope and $b$ is the y-intercept.

• Ordered pair: $(x, y)$ values that satisfy the equation.

Step-by-Step Guidance

1. Choose three values for $x$ (e.g., $x = -2, 0, 2$).

2. Plug each $x$ value into the equation to solve for $y$.

3. Record each $(x, y)$ pair in a table.

4. Plot the points on a coordinate plane and draw a straight line through them.

Try solving on your own before revealing the answer!

Final Answer:

Sample table:

• $x = -2$, $y = -\frac{1}{2}(-2) + 2 = 1 + 2 = 3$ → $(-2, 3)$

• $x = 0$, $y = -\frac{1}{2}(0) + 2 = 0 + 2 = 2$ → $(0, 2)$

• $x = 2$, $y = -\frac{1}{2}(2) + 2 = -1 + 2 = 1$ → $(2, 1)$

Graph: Plot these points and draw a straight line through them.

Q49. For the equation $2x + 3y = 5$, (a) give a table with at least three ordered pairs that are solutions, and (b) graph the equation.

Background

Topic: Linear Equations and Graphing

This question tests your ability to solve for $y$ in terms of $x$ and generate ordered pairs, then graph the equation.

Key terms and formulas:

• Standard form: $Ax + By = C$

• To graph, solve for $y$: $y = \frac{5 - 2x}{3}$

Step-by-Step Guidance

1. Choose three values for $x$ (e.g., $x = 0, 1, 2$).

2. Plug each $x$ value into the equation and solve for $y$.

3. Record each $(x, y)$ pair in a table.

4. Plot the points on a coordinate plane and draw a straight line through them.

Try solving on your own before revealing the answer!

Final Answer:

Sample table:

• $x = 0$, $2(0) + 3y = 5$ → $3y = 5$ → $y = \frac{5}{3}$ → $(0, \frac{5}{3})$

• $x = 1$, $2(1) + 3y = 5$ → $2 + 3y = 5$ → $3y = 3$ → $y = 1$ → $(1, 1)$

• $x = 2$, $2(2) + 3y = 5$ → $4 + 3y = 5$ → $3y = 1$ → $y = \frac{1}{3}$ → $(2, \frac{1}{3})$

Graph: Plot these points and draw a straight line through them.