# Distance Between a Point and a Line In 2D & 3D - Geometry

TLDRThis educational video script offers a detailed guide on calculating the perpendicular distance between a point and a line in both two and three dimensions. It begins by explaining the formula for the distance 'd' in the 2D case, involving the coefficients 'a', 'b', and 'c' of the line equation and the coordinates of the point. The script provides step-by-step calculations using specific examples, including converting equations to standard form and simplifying fractions. It then extends the concept to 3D, introducing an additional variable 'd' for the z-coordinate and demonstrating the calculation with a given point and line equation. The clarity and structure of the script make it an informative resource for understanding geometrical distance calculations.

###### Takeaways

- π The video focuses on teaching how to calculate the perpendicular distance between a point and a line in both two and three dimensions.
- π For a line in 2D with the equation ax + by + c = 0 and a point (x1, y1), the distance formula is |ax1 + by1 + c| / β(aΒ² + bΒ²).
- π An example is given where a line is 3x - 4y + 18 = 0 and a point is (4, 5), resulting in a distance calculation of 2 units.
- π The process involves substituting the coefficients (a, b, c) and the point's coordinates (x1, y1) into the distance formula.
- π The video also covers how to rearrange a given equation into the standard form ax + by + c = 0 for distance calculation purposes.
- π Another example is provided with the equation y = 5x/12 + 1/3, which is rearranged and used to find the distance from the point (3, -6) to the line, resulting in 6 units.
- π The video extends the concept to three dimensions, showing how to calculate the distance from a point (x1, y1, z1) to a line with the equation ax + by + cz + d = 0.
- π An example in 3D is given with the equation y = 1/2x - 1/3z + 2, which is rearranged and used to find the distance from the point (2, 3, -4) to the line, resulting in 4 units.
- π The process for 3D involves multiplying through by a common factor to eliminate fractions and then using the distance formula similar to the 2D case but with an additional term for z.
- π The key takeaway is the methodical approach to finding distances from points to lines in both two and three dimensions using standard form equations and the distance formula.

###### Q & A

### What is the general formula for calculating the distance between a point and a line in two dimensions?

-The general formula to calculate the distance \( d \) between a point \( (x_1, y_1) \) and a line given by \( ax + by + c = 0 \) is \( d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}} \).

### What are the values of a, b, and c in the example line equation '3x - 4y + 18 = 0'?

-In the example line equation '3x - 4y + 18 = 0', the values are \( a = 3 \), \( b = -4 \), and \( c = 18 \).

### What is the point used in the first example of the script?

-The point used in the first example is \( (4, 5) \), where \( x_1 = 4 \) and \( y_1 = 5 \).

### How do you convert the equation 'y = (5/12)x + 1/3' into standard form?

-To convert 'y = (5/12)x + 1/3' into standard form \( ax + by + c = 0 \), multiply through by 12 to eliminate fractions, resulting in \( -5x + 12y + 4 = 0 \).

### What are the values of a and b in the standard form of the equation '5x - 12y + 4 = 0'?

-In the standard form equation '5x - 12y + 4 = 0', the values are \( a = 5 \) and \( b = -12 \).

### What is the point used in the second example of the script?

-The point used in the second example is \( (3, -6) \), where \( x_1 = 3 \) and \( y_1 = -6 \).

### How do you calculate the distance between a point and a line in three dimensions?

-In three dimensions, the distance \( d \) between a point \( (x_1, y_1, z_1) \) and a line given by \( ax + by + cz + d = 0 \) is calculated using \( d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}} \).

### What is the point used in the three-dimensional example of the script?

-The point used in the three-dimensional example is \( (2, 3, -4) \), where \( x_1 = 2 \), \( y_1 = 3 \), and \( z_1 = -4 \).

### What is the rearranged equation for 'y = (1/2)x - (1/3)z + 2' in standard form?

-The rearranged equation in standard form for 'y = (1/2)x - (1/3)z + 2' is \( -\frac{1}{2}x + y + \frac{1}{3}z - 2 = 0 \), which simplifies to \( -3x - 6y + 2z + 12 = 0 \) after multiplying through by 6 to clear fractions.

### What are the values of a, b, c, and d in the standard form of the equation '-3x - 6y + 2z + 12 = 0'?

-In the standard form equation '-3x - 6y + 2z + 12 = 0', the values are \( a = 3 \), \( b = 6 \), \( c = -2 \), and \( d = -12 \).

### How do you ensure the denominator in the distance formula is not zero?

-The denominator in the distance formula \( \sqrt{a^2 + b^2 + c^2} \) (or \( \sqrt{a^2 + b^2} \) in 2D) is the magnitude of the normal vector to the line, which cannot be zero as long as not all coefficients \( a \), \( b \), and \( c \) (or \( a \) and \( b \)) are zero simultaneously.

###### Outlines

##### π Calculating 2D and 3D Distances from Points to Lines

This paragraph introduces the topic of calculating the perpendicular distance from a point to a line in both two and three dimensions. The formula for the distance 'd' in 2D is given as the absolute value of (ax1 + by1 + c) divided by the square root of (a^2 + b^2), where 'a', 'b', and 'c' are the coefficients of the line equation in standard form (ax + by + c = 0), and (x1, y1) are the coordinates of the point. An example is provided using the line equation 3x - 4y + 18 = 0 and the point (4, 5), resulting in a distance of 2 units. The paragraph also begins the process of finding the distance from the point (3, -6) to the graph of the equation y = (5/12)x + (1/3), which is rearranged into standard form for the calculation.

##### π Detailed Calculation of Distance in 2D and Transition to 3D

The second paragraph continues with the calculation of the distance from the point (3, -6) to the line y = (5/12)x + (1/3), which has been converted to the standard form 5x - 12y + 4 = 0. The coefficients are identified as 'a' = 5, 'b' = -12, and 'c' = 4. Using the distance formula, the calculation results in a distance of 6 units. The paragraph then transitions to a three-dimensional problem, where the distance from the point (2, 3, -4) to the graph of y = (1/2)x - (1/3)z + 2 is to be calculated. The equation is rearranged and simplified to -1/2x + y + (1/3)z - 2 = 0, and the coefficients 'a', 'b', 'c', and 'd' are identified as 3, -6, -2, and 12, respectively, setting up for the application of the 3D distance formula.

##### π Completing the 3D Distance Calculation

In the final paragraph, the calculation of the perpendicular distance between the point (2, 3, -4) and the line in 3D is completed. The distance formula for 3D is the absolute value of (ax1 + by1 + cz1 + d) divided by the square root of (a^2 + b^2 + c^2). Plugging in the values, the calculation leads to a distance of 4 units. This concludes the tutorial on how to calculate distances from points to lines in both two and three dimensions.

###### Mindmap

###### Keywords

##### π‘Distance

##### π‘Point

##### π‘Line

##### π‘Equation

##### π‘Standard Form

##### π‘Absolute Value

##### π‘Square Root

##### π‘Two Dimensions

##### π‘Three Dimensions

##### π‘Perpendicular Distance

###### Highlights

Introduction to calculating the distance between a point and a line in two and three dimensions.

Explanation of the generic line equation ax + by + c = 0 for distance calculation.

Presentation of the distance formula d = |ax1 + by1 + c| / β(aΒ² + bΒ²).

Identification of variables a, b, c for the line equation and x1, y1 for the point.

Example calculation with point (4,5) and line equation 3x - 4y + 18 = 0.

Substitution of values into the distance formula with a=3, b=-4, c=18, x1=4, y1=5.

Calculation steps for the distance, including arithmetic operations and square root.

Result of the first example: Distance is 2 units.

Transition to finding distance from point (3, -6) to the graph y = 5x/12 + 1/3.

Conversion of the linear equation to standard form ax + by + c = 0.

Rearrangement and simplification of the equation to 5x - 12y + 4 = 0.

Identification of new variables a=5, b=-12, c=4 for the distance formula.

Second example calculation with point (3, -6) and line equation 5x - 12y + 4 = 0.

Result of the second example: Distance is 6 units.

Introduction to three-dimensional distance calculation with point (2, 3, -4) and line equation y = 1/2x - 1/3z + 2.

Rearrangement of the three-dimensional line equation to standard form.

Identification of three-dimensional variables a=3, b=-6, c=-2, d=12.

Third example calculation with point (2, 3, -4) and line equation 3x - 6y - 2z + 12 = 0.

Final result of the third example: Distance is 4 units.

###### Transcripts

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