- What is the Point-Slope Form
- Point Slope Formula
- Deriving Point-Slope Formula
- Important Notes
- Solved Examples
- Practice Problems
- Frequently Asked Questions

There are majorly three different forms to write the equation of a straight line in a two-dimensional coordinate plane. They are slope-intercept form, general or standard form, and point-slope form. Point slope form is a preferred way to express the equation of a straight line when we know a specific point on the line and its slope. As the name suggests, the point-slope form of a line includes the slope and a point on the line.

The Point-Slope formula is a method used in algebra to express the equation of a straight line. It's particularly helpful when you know a specific point on the line and its slope.

The formula is \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) \) represents the coordinates of the known point, and \( m \) represents the slope of the line.

This formula allows you to find the equation of a line when you have a known point and its slope, aiding in graphing lines and solving various mathematical problems related to linear relationships.

**Example `1`: A line is passing through the point \((2, 5)\) and has a slope of \(3\). Write the equation of the line in point-slope form.**

**Solution:**

To use the point-slope formula, we substitute the given values into the formula \(y - y_1 = m(x - x_1)\). Here, \(m = 3\), \(x_1 = 2\), and \(y_1 = 5\). Plugging these values in, we get:

\(y - 5 = 3(x - 2)\)

So, the equation of the line in point slope form is \(y - 5 = 3(x - 2)\).

**Example 2: Determine the equation of a line passing through the point \((-3, 4)\) with a slope of \(-2\), using point slope form.**

**Solution:**

Applying the point slope formula \(y - y_1 = m(x - x_1)\), where \(m = -2\), \(x_1 = -3\), and \(y_1 = 4\), we have:

\(y - 4 = -2(x - (-3))\)

\(y - 4 = -2(x + 3)\)

Therefore, the equation of the line passing through \((-3, 4)\) with a slope of \(-2\) in point-slope form is \(y - 4 = -2(x + 3)\).

Suppose there is a fixed point \( P_1(x_1, y_1) \) on a non-vertical line \( L \) with a slope of \( m \). Let \( P (x, y) \) represent any arbitrary point on the line \( L \).

The slope of a line passing through the points \( (x_1, y_1) \) and \( (x, y) \) can be defined as the ratio of the difference in `y`-coordinates to the difference in `x`-coordinates:

\( m = \frac{y - y_1}{x - x_1} \)

This equation can be rearranged to isolate \( y - y_1 \), resulting in:

\( y - y_1 = m(x - x_1) \quad \dots \quad (i) \)

Since the point \( P_1(x_1, y_1) \) and all points \( (x, y) \) on \( L \) satisfy equation \( (i) \), it is evident that equation \( (i) \) represents the equation for the given line \( L \).

Therefore, any point \( (x, y) \) lies on the line with slope \( m \) passing through the fixed point \( (x_1, y_1) \) if and only if its coordinates satisfy the equation:

\( y - y_1 = m(x - x_1) \)

Hence, this expression represents the point-slope form of a line equation.

**`1`. Specific Point and Slope:** Point slope form is particularly beneficial when you know a specific point \((x_1, y_1)\) on the line and its slope \(m\). This information allows you to uniquely determine the equation of the line without needing additional data points.

**`2`. Flexible Application:** Point slope form offers flexibility in expressing linear equations. Unlike a slope-intercept form, which requires the `y`-intercept, a point-slope form can be used with any point on the line, making it adaptable to various scenarios where different points are known. Additionally, the point-slope form can be rearranged to write the equation in slope-intercept form and standard form.

**`3`. Straightforward Interpretation:** The point slope form equation \(y - y_1 = m(x - x_1)\) directly relates to the slope \(m\) and the coordinates of the known point \((x_1, y_1)\). This straightforward interpretation makes it easier to understand the geometric and algebraic properties of the line, aiding in graphical representation and problem-solving involving linear relationships.

**Q`1`. A line that passes through the points `(4,7)` and `(5,1)`. Write the equation of the line in point-slope form.**

**Solution:**

To find the equation of the line passing through the points \((4, 7)\) and \((5, 1)\) in point-slope form, we first need to determine the slope (\(m\)) using the formula:

\( m = \frac{y_2 - y_1}{x_2 - x_1} \)

Given that \( (x_1, y_1) = (4, 7) \) and \( (x_2, y_2) = (5, 1) \), we substitute these values into the formula:

\( m = \frac{1 - 7}{5 - 4} \)

\( m = \frac{-6}{1} \)

\( m = -6 \)

Now that we have the slope, we can use one of the given points and the slope to write the equation in point-slope form. Let's use the point \( (4, 7) \):

\( y - y_1 = m(x - x_1) \)

\( y - 7 = -6(x - 4) \)

So, the equation of the line passing through the points \((4, 7)\) and \((5, 1)\) in point-slope form is:

\( y - 7 = -6(x - 4) \)

**Q`2`. The line `y = -2x + 7` passes through the point `(1,5)`. Rewrite the equation of the line in point-slope form.**

**Solution:**

To rewrite the equation of the line \( y = -2x + 7 \) in point-slope form, we need to find the slope (\( m \)) and then use the given point \((1, 5)\) along with the slope to write the equation.

Given that the equation of the line is already in the slope-intercept form \( y = mx + b \), where \( m \) is the slope, we can directly identify the slope as \( m = -2 \).

Now, using the point-slope form formula:

\( y - y_1 = m(x - x_1) \)

We substitute \( (x_1, y_1) = (1, 5) \) and \( m = -2 \) into the equation:

\( y - 5 = -2(x - 1) \)

So, the equation of the line \( y = -2x + 7 \) rewritten in point-slope form, considering it passes through the point \((1, 5)\), is:

\( y - 5 = -2(x - 1) \)

**Q`3`. A line is perpendicular to `y = -3x + 4` and passes through `(6, -3)`. Write the equation of the line in point-slope form.**

**Solution:**

To find the equation of the line perpendicular to \( y = -3x + 4 \) and passing through the point \( (6, -3) \) in point-slope form, we first need to determine the slope of the given line and then find the negative reciprocal of that slope to get the slope of the perpendicular line.

The slope of the given line \( y = -3x + 4 \) is \( m_1 = -3 \).

The negative reciprocal of \( m_1 \) is \( m_2 = \frac{1}{3} \).

Now, using the point-slope form formula:

\( y - y_1 = m(x - x_1) \)

We substitute \( (x_1, y_1) = (6, -3) \) and \( m_2 = \frac{1}{3} \) into the equation:

\( y - (-3) = \frac{1}{3}(x - 6) \)

Simplifying:

\( y + 3 = \frac{1}{3}x - 2 \)

\( y = \frac{1}{3}x - 2 - 3 \)

\( y = \frac{1}{3}x - 5 \)

So, the equation of the line perpendicular to \( y = -3x + 4 \) and passing through the point \( (6, -3) \) in point-slope form is:

\( y = \frac{1}{3}x - 5 \)

**Q`4`. The equation of a line in point-slope form is `y + 4 = -2(x - 4)`. Write the equation of the line in the slope-intercept form.**

**Solution:**

To write the equation of the line given in point-slope form \(y + 4 = -2(x - 4)\) into slope-intercept form \(y = mx + b\), we need to isolate \(y\) on one side of the equation.

First, let's distribute \(-2\) into \((x - 4)\):

\(y + 4 = -2x + 8\)

Now, subtract \(4\) from both sides to isolate \(y\):

\(y = -2x + 8 - 4\)

\(y = -2x + 4\)

So, the equation of the line in slope-intercept form is:

\(y = -2x + 4\)

**Q`5`. A line is parallel to `y = 1/2x -5` and passes through `(-6, 4)`. Write the equation of the line in point-slope form**

**Solution:**

To write the equation of the line parallel to \( y = \frac{1}{2}x - 5 \) and passing through the point \( (-6, 4) \) in point-slope form, we first need to determine the slope of the given line, and then we can use that slope along with the given point to write the equation.

The slope of the given line \( y = \frac{1}{2}x - 5 \) is \( m_1 = \frac{1}{2} \).

Since the line we're trying to find is parallel to this line, it will have the same slope. So, the slope of our line is also \( m_2 = \frac{1}{2} \).

Now, using the point-slope form formula:

\( y - y_1 = m(x - x_1) \)

We substitute \( (x_1, y_1) = (-6, 4) \) and \( m_2 = \frac{1}{2} \) into the equation:

\( y - 4 = \frac{1}{2}(x + 6) \)

Hence, the equation of the line in the point-slope form:

\( y - 4 = \frac{1}{2}(x + 6) \)

**Q`1`. Which of the following represents the equation of a line passing through the point \((-1, 3)\) with a slope of \(4\) in point slope form?**

- \(y + 3 = 4(x - (-1))\)
- \(y - 3 = 4(x - (-1))\)
- \(y - 3 = 4(x - 1)\)
- \(y - 3 = -4(x + 1)\)

**Answer:** b.

**Q`2`. What is the point-slope from the equation of a line with a slope of \(\frac{1}{2}\) passing through the point \((2, 5)\)?**

- \(y - 5 = \frac{1}{2}(x - 2)\)
- \(y + 5 = \frac{1}{2}(x - 2)\)
- \(y - 5 = \frac{1}{2}(x + 2)\)
- \(y + 5 = \frac{1}{2}(x + 2)\)

**Answer:** a.

**Q`3`. Which of the following equations is written in point-slope form?**

- \(y = 2x - 3\)
- \(2y - 4x = 8\)
- \(y - 4 = 2(x - 1)\)
- \(3x + 5y = 15\)

**Answer:** c.

**Q`4`. A line is perpendicular to \( y = -3x + 4 \) and passes through the point \( (6, -3) \). Write the equation of the line in point-slope form.**

- \( y + 3 = 3(x - 6) \)
- \( y + 3 = \frac{1}{3}(x - 6) \)
- \( y - 3 = -\frac{1}{3}(x - 6) \)
- \( y - 3 = 3(x - 6) \)

**Answer:** b.

**Q`5`. A line is parallel to \( y = \frac{1}{2}x - 5 \) and passes through the point \( (-6, 4) \). Write the equation of the line in point-slope form.**

- \( y - 4 = \frac{1}{2}(x + 6) \)
- \( y + 4 = -2(x + 6) \)
- \( y - 4 = -\frac{1}{2}(x + 6) \)
- \( y + 4 = \frac{1}{2}(x - 6) \)

**Answer:** a.

**Q`1`: What is the point-slope form used for?**

**Answer:** Point slope form is used to express the equation of a straight line when you know a specific point on the line and its slope. It's particularly useful in mathematics for graphing lines and solving problems involving linear relationships.

**Q`2`: How do you convert an equation from point slope form to slope-intercept form?**

**Answer:** To convert an equation from point slope form \(y - y_1 = m(x - x_1)\) to slope-intercept form \(y = mx + b\), you simply need to isolate \(y\) by adding \(y_1\) to both sides and then simplifying.

**Q`3`: Can any point on the line be used to write the equation of the line in point-slope form?**

**Answer:** Yes, point slope form allows you to use any point on the line to express the equation. However, it's typically most convenient to use a lattice point to make calculations easier.

**Q`4`: What is the point-slope form of a line whose slope is zero or undefined?**

**Answer:** If the slope of the line is zero, the equation in point-slope form becomes \(y - y_1 = 0(x - x_1)\), simplifying to \(y = y_1\), which represents a horizontal line. If the slope is undefined (vertical line), the equation cannot be expressed in point-slope form as it does not have a slope \(m\). A vertical line is written as \(x = x_1\)

**Q`5`: When is the point-slope form preferred over the slope-intercept form?**

**Answer:** Point slope form is preferred when you know any specific point on the line and its slope. It's particularly useful in situations where you have the coordinates of any point on the line and the slope of the line, making it easier to write the equation directly without needing to calculate the `y`-intercept.