Intermediate Algebra Problems with Solutions:
Lines, Functions, and Triangles

Problem: Line \( L \) passes through the points \( (1, 3) \) and \( (-3, 4) \).

1. Find the slope of line \( L \).
2. Find the y-intercept of \( L \).
3. Find the x-intercept of \( L \).
4. Find \( b \) so that line \( L \) passes through the point \( (-4, b) \).

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Solution:

1. Slope: \[ \text{slope} = \dfrac{4 - 3}{-3 - 1} = -\dfrac{1}{4} \]
2. Equation of line \( L \) (point-slope form): \[ y - 3 = -\dfrac{1}{4}(x - 1) \] To find the y-intercept, let \( x = 0 \): \[ y - 3 = -\dfrac{1}{4}(0 - 1) = \dfrac{1}{4} \implies y = \dfrac{13}{4} \]
3. To find the x-intercept, let \( y = 0 \): \[ 0 - 3 = -\dfrac{1}{4}(x - 1) \implies -3 = -\dfrac{1}{4}(x - 1) \] Multiply both sides by 4 and solve for \( x \): \[ -12 = -(x - 1) \implies x - 1 = 12 \implies x = 13 \]
4. Substitute \( x = -4 \) and \( y = b \) into the equation: \[ b - 3 = -\dfrac{1}{4}(-4 - 1) = -\dfrac{1}{4}(-5) = \dfrac{5}{4} \implies b = 3 + \dfrac{5}{4} = \dfrac{17}{4} \]

Problem: The graph of function \( f \) is shown below.

1. Find the domain of \( f \).
2. Find the range of \( f \).
3. Find \( f(0) \), \( f(1) \), and \( f(2) \).
4. Find all values of \( x \) for which \( f(x) = 5 \).

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Solution:

1. Domain: \( (-4, 4] \) (interval open at \( x = -4 \) because the function is undefined at this value of \( x \)).
2. Range: \( [-4, 12] \)
3. \( f(0) = -4 \), \( f(1) = -3 \), \( f(2) = 0 \)
4. \( f(x) = 5 \) for \( x = -3 \) or \( x = 3 \)

Problem: The graph of line segment \( AB \) is shown below.

1. Use the graph to find the coordinates of points \( A \) and \( B \).
2. Find the length of the line segment \( AB \).
3. Find the midpoint of the line segment \( AB \).

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Solution:

1. \( A(-2, 4) \), \( B(4, 2) \)
2. Length of \( AB = \sqrt{(4 - (-2))^2 + (2 - 4)^2} = \sqrt{6^2 + (-2)^2} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10} \)
3. Midpoint coordinates: \( x = \dfrac{-2 + 4}{2} = 1 \), \( y = \dfrac{4 + 2}{2} = 3 \)

Problem: Line \( L_1 \) passes through the point \( A(1, b) \) as shown below. Find the equation of line \( L_2 \) that passes through point \( A \) and is perpendicular to line \( L_1 \).

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Solution:

From the graph, the x-intercept is \( (-3, 0) \) and the y-intercept is \( (0, 2) \).

Slope of \( L_1 \):

\[ \text{Slope of } L_1 = \dfrac{2 - 0}{0 - (-3)} = \dfrac{2}{3} \]

Find \( b \) using the slope, point \( (1, b) \), and point \( (0, 2) \):

\[ \dfrac{b - 2}{1 - 0} = \dfrac{2}{3} \implies b - 2 = \dfrac{2}{3} \implies b = \dfrac{8}{3} \]

Slope of \( L_2 \) (perpendicular to \( L_1 \)):

\[ \text{Slope of } L_2 = -\dfrac{1}{\dfrac{2}{3}} = -\dfrac{3}{2} \]

Equation of \( L_2 \) using point-slope form at \( (1, \dfrac{8}{3}) \):

\[ y - \dfrac{8}{3} = -\dfrac{3}{2}(x - 1) \]

Simplified:

\[ y = -\dfrac{3}{2}x + \dfrac{25}{6} \]

Problem: Are the two lines \( L_1 \) through the points \( (-1, 4) \) and \( (0, 4) \), and \( L_2 \) through the points \( (5, 6) \) and \( (5, 2) \), parallel, perpendicular, or neither? Explain your answer.

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Solution:

Slope of \( L_1 = \dfrac{4 - 4}{0 - (-1)} = 0 \), so \( L_1 \) is horizontal.
Slope of \( L_2 = \dfrac{2 - 6}{5 - 5} = \dfrac{-4}{0} \), which is undefined, so \( L_2 \) is vertical.
Therefore, the two lines are perpendicular.

Problem: Find all values of the constant \( K \) so that the inequality \( 5(x - K + 2) \geq 2(x + 4) - 7 \) has a solution set given by the interval \( [10, +\infty) \).

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Solution:

Solve the given inequality:

\[ 5(x - K + 2) \geq 2(x + 4) - 7 \] \[ 5x - 5K + 10 \geq 2x + 8 - 7 \] \[ 3x \geq 5K - 9 \implies x \geq -3 + \dfrac{5K}{3} \]

For the solution set to be equal to \( [10, +\infty) \), we set:

\[ -3 + \dfrac{5K}{3} = 10 \implies \dfrac{5K}{3} = 13 \implies 5K = 39 \implies K = \dfrac{39}{5} \]

Problem: Linda walked for \( 2 \) hours then ran for \( 1 \) hour. If she runs three times as fast as she walks and the total trip was \( 20 \) kilometers, then what is the running speed of Linda?

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Solution:

Let \( x \) be the walking speed (in km/hr). Then, the running speed is \( 3x \).

The distance walked is \( 2 \times x = 2x \) km, and the distance run is \( 1 \times 3x = 3x \) km.

The total distance is 20 km. Hence:

\[ 2x + 3x = 20 \implies 5x = 20 \implies x = 4\text{ km/hr} \]

Therefore, the running speed of Linda is:

\[ 3x = 3 \times 4 = 12\text{ km/hr} \]

Problem: Each pair of the three lines defined by their equations \( y = \dfrac{1}{2}x - 1 \), \( y = 2x + 2 \), and \( y = -x + 2 \) has a point of intersection so that when put together there are three points of intersection making a triangle \( ABC \). Show that triangle \( ABC \) is isosceles.

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Solution:

Let \( A \) be the point of intersection of \( y = \dfrac{1}{2}x - 1 \) and \( y = 2x + 2 \):

\[ \dfrac{1}{2}x - 1 = 2x + 2 \implies -\dfrac{3}{2}x = 3 \implies x = -2 \] \[ y = 2(-2) + 2 = -2 \implies A = (-2, -2) \]

Let \( B \) be the point of intersection of \( y = \dfrac{1}{2}x - 1 \) and \( y = -x + 2 \):

\[ \dfrac{1}{2}x - 1 = -x + 2 \implies \dfrac{3}{2}x = 3 \implies x = 2 \] \[ y = \dfrac{1}{2}(2) - 1 = 0 \implies B = (2, 0) \]

Let \( C \) be the point of intersection of \( y = -x + 2 \) and \( y = 2x + 2 \):

\[ -x + 2 = 2x + 2 \implies -3x = 0 \implies x = 0 \] \[ y = 2(0) + 2 = 2 \implies C = (0, 2) \]

Using the distance formula \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \):

• Length of segment \( AB \): \[ AB = \sqrt{(2 - (-2))^2 + (0 - (-2))^2} = \sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5} \]
• Length of segment \( AC \): \[ AC = \sqrt{(0 - (-2))^2 + (2 - (-2))^2} = \sqrt{2^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \]
• Length of segment \( BC \): \[ BC = \sqrt{(0 - 2)^2 + (2 - 0)^2} = \sqrt{(-2)^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \]

Since \( AB = AC = 2\sqrt{5} \), triangle \( ABC \) is isosceles.