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

This collection of intermediate algebra problems covers key topics such as lines, slope, functions, domain and range, and x- and y-intercepts. Each problem includes detailed solutions and clear explanations to help deepen your understanding.

Questions 1

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

  1. Fine 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 \( (-4,b) \).

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 \( x = 0 \) in the equation of the line: \[ y - 3 = - \dfrac{1}{4}(0 - 1) = \dfrac{1}{4} \Rightarrow y = \dfrac{13}{4} \]
  3. To find the x-intercept, let \( y = 0 \) in the equation of the line: \[ 0 - 3 = - \dfrac{1}{4}(x - 1) \Rightarrow -3 = \dfrac{-1}{4}(x - 1) \] Multiply both sides by 4 and solve for \( x \): \[ -12 = -(x - 1) \Rightarrow x - 1 = 12 \Rightarrow x = 13 \]
  4. Substitute \( x = -4 \) and \( y = b \) into the equation: \[ b - 3 = \dfrac{-1}{4}(-4 - 1) = \dfrac{-1}{4}(-5) = \dfrac{5}{4} \Rightarrow b = \dfrac{17}{4} \]

Questions 2

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 \( f(x) = 5 \).

intermediate algebra problem 2

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 \( x = 3 \)

Questions 3

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 \( AB \).

intermediate algebra problem 3

Solution:

  1. A\((-2 , 4)\), B\((4 , 2)\)

  2. Length of \( AB = \sqrt{(4 - (-2))^2 + (2 - 4)^2} = 2\sqrt{10} \)

  3. Midpoint coordinates: \( x = \dfrac{-2 + 4}{2} = 1 \), \( y = \dfrac{4 + 2}{2} = 3 \)

Questions 4

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 \( L_1 \).

intermediate algebra problem 4

Solution:

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

Slope of L1: \[ \text{Slope of } L_1 = \dfrac{2 - 0}{0 - (-3)} = \dfrac{2}{3} \] Find \( b \): using the slope, point \((1, b)\) and point \( (0,2) \) \[ \dfrac{b - 2}{1 - 0} = \dfrac{2}{3} \] Solve: \[ b - 2 = \dfrac{2}{3} \Rightarrow b = \dfrac{8}{3} \] Slope of L2 (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: \[ y = -\dfrac{3}{2}x + \dfrac{25}{6} \]

Questions 5

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.

Solution:

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

Questions 6

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 \([10, +\infty)\).

Solution:

Solve the given inequality \[ 5(x - K + 2) \geq 2(x + 4) - 7 \] to obtain the solution set \[ x \geq -3 + \dfrac{5K}{3} \] For the above solution set to be equal to the set \([10, +\infty)\), we need to have \[ -3 + \dfrac{5K}{3} = 10 \] Solving the above for \(K\) gives \[ K = \dfrac{39}{5} \]

Questions 7

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?

Solution:

Let \( x \) be the walking speed (in km/hr). Then, the running speed is \( 3x \). The distance walked is \( 2 \times x = 2x \) kilometers, and the distance run is \( 1 \times 3x = 3x \) kilometers. The total distance is 20 km. Hence, we have the equation: \[ 2x + 3x = 20 \] Solving for \( x \): \[ 5x = 20 \implies x = 4 \text{ km/hr} \] Therefore, the running speed of Linda is: \[ 3x = 3 \times 4 = 12 \text{ km/hr} \]

Questions 8

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.

Solution:

Let \( A \) the be the point of intersection of the lines: \[ y = \dfrac{1}{2}x - 1 \quad \text{and} \quad y = 2x + 2. \] Solve the system of equations formed by these two lines to find \( A \). \[ \dfrac{1}{2}x - 1 = 2x + 2 \] Solve for \( x \): \[ \dfrac{1}{2}x - 1 = 2x + 2 \implies \dfrac{1}{2}x - 2x = 2 + 1 \implies -\dfrac{3}{2}x = 3 \implies x = -2. \] Substitute \( x = -2 \) into one of the equations: \[ y = 2(-2) + 2 = -4 + 2 = -2. \] Hence \[A = (-2, -2). \] Let \( B \) the be the point of intersection of the lines: \[ y = \dfrac{1}{2}x - 1 \quad \text{and} \quad y = - x + 2. \] Solve the system of equations formed by these two lines to find \( B \). \[ \dfrac{1}{2}x - 1 = -x + 2 \] \[ \dfrac{1}{2}x + x = 2 + 1 \implies \dfrac{3}{2}x = 3 \implies x = 2. \] \[ y = \dfrac{1}{2}(2) - 1 = 1 - 1 = 0. \] \[ B = (2, 0). \] Let \( C \) the be the point of intersection of the lines: \[ y = -x + 2 \quad \text{and} \quad y = 2x + 2. \] Solve the system of equations formed by these two lines to find \( C \). \[ -x + 2 = 2x + 2 \implies -x - 2x = 2 - 2 \implies -3x = 0 \implies x = 0. \] \[ y = 2(0) + 2 = 2. \] \[ C = (0, 2). \] The Distance formula for points \( (x_1, y_1) \text{ and } (x_2, y_2) \) is given by: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}. \] 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 \( AB = AC = 2\sqrt{5} \), the triangle \( ABC \) is isosceles.

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