Equations of Lines: Practice Problems with Step-by-Step Solutions

Question: Find the equation of the line that passes through the points \((-1 , 0)\) and \((-4 , 12)\).

---

Solution:

First, find the slope \(m\) of the line:

\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{12 - 0}{-4 - (-1)} = \frac{12}{-3} = -4 \]

Next, use the point-slope form \(y - y_1 = m (x - x_1)\) with the point \((-1, 0)\):

\[ y - 0 = -4(x - (-1)) \]

Simplify and write the equation in general form:

\[ y = -4(x + 1) \] \[ y = -4x - 4 \] \[ 4x + y = -4 \]

Question: What is the equation of the line passing through the points \((-2 , 0)\) and \((-2 , 4)\)?

---

Solution:

Notice that the two points have the exact same \(x\)-coordinate. This means they lie on the same vertical line. The equation of a vertical line is \(x = c\), where \(c\) is the constant \(x\)-coordinate of every point on the line.

\[ x = -2 \]

Question: Find the equation of the line that passes through the points \((7 , 5)\) and \((-9 , 5)\).

---

Solution:

Notice that the two points have the same \(y\)-coordinate. This indicates they lie on a horizontal line. The equation of a horizontal line takes the form \(y = c\).

\[ y = 5 \]

Question: Find the equation of the line through the point \((3 , 4)\) that is parallel to the \(x\)-axis.

---

Solution:

A line parallel to the \(x\)-axis is a horizontal line, so it has a slope of \(0\) and its equation is of the form \(y = \text{constant}\). Since it passes through \((3 , 4)\), the \(y\)-coordinate must be \(4\).

\[ y = 4 \]

Question: What is the equation of the line through the point \((-3 , 2)\) that has an \(x\)-intercept at \(x = -1\)?

---

Solution:

An \(x\)-intercept at \(x = -1\) gives a second point on the line: \((-1, 0)\). Now, find the slope of the line:

\[ m = \frac{2 - 0}{-3 - (-1)} = \frac{2}{-2} = -1 \]

Using the point-slope form with the intercept \((-1, 0)\):

\[ y - 0 = -1 (x - (-1)) \]

Simplify into slope-intercept form:

\[ y = -x - 1 \]

Question: Find the equation of the line that has an \(x\)-intercept at \(x = -4\) and a \(y\)-intercept at \(y = 5\).

---

Solution:

The \(x\)- and \(y\)-intercepts correspond to the points \((-4, 0)\) and \((0, 5)\). First, find the slope:

\[ m = \frac{5 - 0}{0 - (-4)} = \frac{5}{4} \]

Since we already know the \(y\)-intercept \((b = 5)\), write the slope-intercept form directly:

\[ y = \frac{5}{4}x + 5 \]

Alternatively, write it in general form by multiplying all terms by \(4\) to remove fractions:

\[ 4y = 5x + 20 \] \[ 5x - 4y = -20 \]

Question: What is the equation of the line through the point \((-1 , 0)\) that is perpendicular to the line \(y = 9\)?

---

Solution:

The line \(y = 9\) is horizontal. A line perpendicular to a horizontal line must be vertical, which takes the form \(x = \text{constant}\). Because our line passes through \((-1 , 0)\), the constant is the \(x\)-coordinate.

\[ x = -1 \]

Question: Find the slope, the \(x\)-intercept, and the \(y\)-intercept of the line given by the equation \(-3x + 5y = 8\).

---

Solution:

To find the slope, rewrite the equation in slope-intercept form \(y = mx + b\):

\[ 5y = 3x + 8 \] \[ y = \frac{3}{5}x + \frac{8}{5} \]

The slope is \(m = \dfrac{3}{5}\). The \(y\)-intercept is \(y = \dfrac{8}{5}\).

To find the \(x\)-intercept, set \(y = 0\) in the original equation and solve for \(x\):

\[ -3x + 5(0) = 8 \] \[ -3x = 8 \] \[ x = -\frac{8}{3} \]

Question: Find the slope-intercept form for the line given by its equation \(\dfrac{x}{4} - \dfrac{y}{5} = 3\).

---

Solution:

Isolate the \(y\) term to convert the equation into \(y = mx + b\) form:

\[ -\frac{y}{5} = -\frac{x}{4} + 3 \]

Multiply the entire equation by \(-5\):

\[ y = \frac{5}{4}x - 15 \]

Question: Are the lines \(x = -3\) and \(x = 0\) parallel or perpendicular?

---

Solution:

The equation \(x = -3\) represents a vertical line. The equation \(x = 0\) is the \(y\)-axis, which is also a vertical line. Since both lines are vertical and have different \(x\)-intercepts, they never intersect.

Answer: The two lines are parallel.

Question: For what value of \(b\) is the point \((2 , 2b)\) on the line with equation \(x - 4y = 6\)?

---

Solution:

For a point to lie on a line, its coordinates must satisfy the equation. Substitute \(x = 2\) and \(y = 2b\) into the equation:

\[ (2) - 4(2b) = 6 \] \[ 2 - 8b = 6 \]

Solve for \(b\):

\[ -8b = 4 \] \[ b = -\frac{4}{8} \] \[ b = -\frac{1}{2} \]