Master Intermediate Algebra: Equations of Lines
Practice Problems & Detailed Solutions

Problem: Find the equation of the line that passes through the point \( (0, -6) \) and has a slope equal to \( 5 \).

---

Solution:

Given a slope of \( 5 \) and the y-intercept at \( (0, -6) \), the slope-intercept form of the equation is: \[ y = 5x - 6 \]

Problem: Find the equation of the line that passes through the point \( (-2, 3) \) and has a slope equal to \( -2 \).

---

Solution:

Given a slope of \( -2 \) and the point \( (-2, 3) \), the point-slope form of the equation is: \[ y - 3 = -2(x - (-2)) \] \[ y - 3 = -2(x + 2) \] In slope-intercept form: \[ y = -2x - 1 \]

Problem: Is there a line that passes through all the following points: \( A(2, 3) \), \( B(4, 7) \), and \( C(5, 6) \)?

---

Solution:

For a line to pass through all three points, the points must be collinear.

Find the slope through \( AB \):

\[ m_{AB} = \dfrac{7 - 3}{4 - 2} = 2 \]

Find the slope through \( BC \):

\[ m_{BC} = \dfrac{6 - 7}{5 - 4} = -1 \]

Since the slopes through \( AB \) and \( BC \) are not equal (\( 2 \neq -1 \)), points \( A \), \( B \), and \( C \) are not collinear, and no single line passes through all three points.

Problem: Find the equation of the line through the point \( (3, 0) \) that is perpendicular to the line passing through the points \( (2, 1) \) and \( (4, 5) \).

---

Solution:

First, find the slope \( m \) of the line through \( (2, 1) \) and \( (4, 5) \):

\[ m = \dfrac{5 - 1}{4 - 2} = 2 \]

The slope \( s \) of the line perpendicular to a line of slope \( 2 \) satisfies \( m \times s = -1 \):

\[ 2 \times s = -1 \implies s = -\dfrac{1}{2} \]

Using the point-slope form with point \( (3, 0) \) and slope \( -\dfrac{1}{2} \):

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

Problem: What are the values of \( m \) and \( w \) so that the line with equation \( y - mx = w \) passes through the points \( (0, 4) \) and \( (2, 0) \)?

---

Solution:

Substitute the point \( (0, 4) \) (\( x = 0, y = 4 \)) into the equation:

\[ 4 - m(0) = w \implies w = 4 \]

Substitute the point \( (2, 0) \) (\( x = 2, y = 0 \)) and \( w = 4 \) into the equation:

\[ 0 - m(2) = 4 \implies -2m = 4 \implies m = -2 \]

Therefore, \( m = -2 \) and \( w = 4 \).

Problem: Find the equation of the line through the point \( (-4, 5) \) that is perpendicular to the x-axis.

---

Solution:

A line perpendicular to the x-axis is a vertical line parallel to the y-axis, having the general equation \( x = \text{constant} \).

Passing through \( (-4, 5) \), the equation is:

\[ x = -4 \]

Problem: Find the equation of the line through the point \( (6, -1) \) that is perpendicular to the y-axis.

---

Solution:

A line perpendicular to the y-axis is a horizontal line parallel to the x-axis, having the general equation \( y = \text{constant} \).

Passing through \( (6, -1) \), the equation is:

\[ y = -1 \]

Problem: The graph shown below is that of a linear function \( f \). Use the graph to approximate \( f(120) \).

---

Solution:

The function \( f \) is linear and passes through the origin, so it has the form \( f(x) = mx \).

Using the points \( (0, 0) \) and \( (4, 1) \) from the graph, find the slope \( m \):

\[ m = \dfrac{1 - 0}{4 - 0} = \dfrac{1}{4} \]

Now calculate \( f(120) \):

\[ f(120) = \dfrac{1}{4}(120) = 30 \]

Problem: Is the line with equation \( 0.5y + 0.5x = 5 \) perpendicular to the line shown in the graph for Question 8? Explain your answer.

---

Solution:

Write the equation \( 0.5y + 0.5x = 5 \) in slope-intercept form:

\[ 0.5y = -0.5x + 5 \implies y = -x + 10 \]

The slope of this line is \( m_1 = -1 \).

From Question 8, the slope of the graphed line is \( m_2 = \dfrac{1}{4} \).

For two lines to be perpendicular, the product of their slopes must be \( -1 \):

\[ m_1 \times m_2 = (-1)\left(\dfrac{1}{4}\right) = -\dfrac{1}{4} \neq -1 \]

Therefore, the line with equation \( 0.5y + 0.5x = 5 \) is not perpendicular to the line given by the graph.

Problem: If the two lines with equations \( 2x - 3y = 5 \) and \( -4x + 5y = 7 \) are graphed in the same coordinate system, will they have a point of intersection? If yes, what are its coordinates?

---

Solution:

Solve the system of linear equations:

\[ \begin{cases} 2x - 3y = 5 \\ -4x + 5y = 7 \end{cases} \]

Multiply the first equation by 2:

\[ 4x - 6y = 10 \]

Add this to the second equation:

\[ (4x - 6y) + (-4x + 5y) = 10 + 7 \implies -y = 17 \implies y = -17 \]

Substitute \( y = -17 \) into \( 2x - 3y = 5 \):

\[ 2x - 3(-17) = 5 \implies 2x + 51 = 5 \implies 2x = -46 \implies x = -23 \]

The two lines intersect at the point: \( (-23, -17) \)

Problem: Find the equations of two perpendicular lines with a point of intersection at \( (2, -3) \), where one of the lines passes through the origin.

---

Solution:

The first line passes through the origin \( (0, 0) \) and \( (2, -3) \). Its slope \( m_1 \) is:

\[ m_1 = \dfrac{-3 - 0}{2 - 0} = -\dfrac{3}{2} \]

Its equation is:

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

The second line is perpendicular to the first, so its slope \( m_2 \) satisfies \( -\dfrac{3}{2} \times m_2 = -1 \):

\[ m_2 = \dfrac{2}{3} \]

Using the point-slope form with point \( (2, -3) \) and slope \( \dfrac{2}{3} \):

\[ y - (-3) = \dfrac{2}{3}(x - 2) \implies y + 3 = \dfrac{2}{3}x - \dfrac{4}{3} \implies y = \dfrac{2}{3}x - \dfrac{13}{3} \]

Problem: Find the equations of the lines that pass through the origin and are tangent to the circle with center at \( (2, 4) \) and radius equal to \( 1 \).

---

Solution:

Step 1: Equation of the Line
A line passing through the origin has the form \( y = kx \).

Step 2: Equation of the Circle
The equation of the circle with center \( (2, 4) \) and radius \( 1 \) is: \[ (x - 2)^2 + (y - 4)^2 = 1 \]

Step 3: Substitute \( y = kx \) into the Circle Equation
\[ (x - 2)^2 + (kx - 4)^2 = 1 \] \[ x^2 - 4x + 4 + k^2x^2 - 8kx + 16 = 1 \] \[ (1 + k^2)x^2 + (-4 - 8k)x + 19 = 0 \]

Step 4: Tangent Condition
For the line to be tangent to the circle, the quadratic equation must have exactly one real solution, meaning the discriminant \( \Delta \) must be zero: \[ \Delta = (-4 - 8k)^2 - 4(1 + k^2)(19) = 0 \] \[ 16 + 64k + 64k^2 - 76 - 76k^2 = 0 \] \[ -12k^2 + 64k - 60 = 0 \] Divide by \( -4 \): \[ 3k^2 - 16k + 15 = 0 \]

Step 5: Solve for \( k \)
Using the quadratic formula: \[ k = \dfrac{16 \pm \sqrt{(-16)^2 - 4(3)(15)}}{2(3)} = \dfrac{16 \pm \sqrt{256 - 180}}{6} = \dfrac{16 \pm \sqrt{76}}{6} \] \[ k = \dfrac{16 \pm 2\sqrt{19}}{6} = \dfrac{8 \pm \sqrt{19}}{3} \]

Step 6: Final Line Equations
\[ y = \left(\dfrac{8 + \sqrt{19}}{3}\right)x \quad \text{and} \quad y = \left(\dfrac{8 - \sqrt{19}}{3}\right)x \]

Problem: Which of the points \( A(3, 0) \), \( B(-2, -5) \), and \( C(0, 2) \) lie on the graph of the line whose equation is given by \( -3y + 2x - 6 = 0 \)?

---

Solution:

For a point to lie on the graph of a line, its coordinates must satisfy the equation of the line.

Check point \( A(3, 0) \):
\[ -3(0) + 2(3) - 6 = 0 + 6 - 6 = 0 \] The equation is satisfied, so point \( A \) lies on the line.

Check point \( B(-2, -5) \):
\[ -3(-5) + 2(-2) - 6 = 15 - 4 - 6 = 5 \neq 0 \] The equation is not satisfied, so point \( B \) does not lie on the line.

Check point \( C(0, 2) \):
\[ -3(2) + 2(0) - 6 = -6 + 0 - 6 = -12 \neq 0 \] The equation is not satisfied, so point \( C \) does not lie on the line.