Physics and Mathematics
Perpendicular Distance of a Point from a Given Line
1. Statement of the Concept
The perpendicular distance of a point from a given line is the shortest distance between the point and the line.
This shortest distance is always measured along the perpendicular drawn from the point to the line.
2. Explanation and Mathematical Derivation
Case: Distance of Point [(x_1, y_1)] from Line [Ax + By + C = 0]
Step 1: General Equation of Line
Let the given line be:
[Ax + By + C = 0]
Let the given point be:
[(x_1, y_1)]
Step 2: Concept of Normal Direction
- The coefficients [A] and [B] act as direction ratios of the normal to the line.
- The shortest distance from the point to the line is measured along this normal.
Step 3: Distance Formula
The perpendicular distance [d] of the point from the line is given by:
[d = \dfrac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}]
Why Absolute Value?
- Distance is always non-negative
- The expression [Ax_1 + By_1 + C] may be positive or negative depending on the side of the line
3. Key Features and Important Observations
- Formula works only when the line is in general form
[Ax + By + C = 0] - Distance is independent of orientation of axes
- If point lies on the line ⇒ distance = 0
- Absolute value ensures distance, not displacement
- Same formula is valid for:
- distance from origin
- distance between parallel lines (with modification)
4. Important Formulas to Remember
| Situation | Formula |
|---|---|
| Distance of [(x_1, y_1)] from [Ax + By + C = 0] | [\dfrac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}] |
| Distance from origin | [\dfrac{|C|}{\sqrt{A^2 + B^2}}] |
| Point lies on line | Distance = 0 |
5. Conceptual Questions with Solutions
1. Why is the perpendicular distance the shortest distance?
Among all possible segments joining a point to a line, the perpendicular segment minimizes length.
This follows from triangle inequality and basic geometry: the perpendicular forms the shortest side.
2. Why do coefficients [A] and [B] represent the normal direction?
For the line [Ax + By + C = 0], the vector [(A, B)] is perpendicular to the line because it is orthogonal to every direction vector of the line.
Hence distance is measured along this direction.
3. What does [Ax_1 + By_1 + C] physically represent?
It measures the signed projection of the point onto the normal direction.
Sign tells on which side of the line the point lies; magnitude gives distance after normalization.
4. Can distance ever be negative?
No.
Distance is a scalar quantity.
Negative sign only indicates direction, which is removed by absolute value.
5. Why do we divide by [\sqrt{A^2 + B^2}]?
Because [(A, B)] is not a unit normal vector.
Dividing by its magnitude converts projection into actual distance.
7. FAQ / Common Misconceptions
1. Students forget to convert the line into general form.
The formula is valid only for [Ax + By + C = 0].
All terms must be on one side.
2. Absolute value is often ignored.
Ignoring modulus may give a negative answer, which is meaningless for distance.
3. Dividing only by [A + B] instead of [\sqrt{A^2 + B^2}].
This is a conceptual error.
Normalization requires magnitude, not sum.
4. Confusion between distance from origin and distance from any point.
Distance from origin is just a special case where [(x_1, y_1) = (0, 0)].
5. Belief that distance depends on slope.
Distance depends on position, not inclination.
8. Practice Questions with Step-by-Step Solutions
Question 1: Find the perpendicular distance of the point [(3, 4)] from the line [3x + 4y − 10 = 0].
Step-by-Step Solution:
1. Given line: [A = 3, B = 4, C = −10]
2. Given point: [(x_1, y_1) = (3, 4)]
3. Apply distance formula:
[d = \dfrac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}]
4. Substitute values:
[d = \dfrac{|3(3) + 4(4) − 10|}{\sqrt{3^2 + 4^2}}]
5. Simplify numerator:
[
|9 + 16 − 10| = |15| = 15
]
6. Simplify denominator:
[
\sqrt{9 + 16} = \sqrt{25} = 5
]
Answer:
[d = 3]
Question 2: Find the perpendicular distance of the origin from the line [5x − 12y + 13 = 0].
Step-by-Step Solution:
1. Point: [(0, 0)]
2. Use distance from origin:
[
d = \dfrac{|C|}{\sqrt{A^2 + B^2}}
]
3. Substitute values:
[
d = \dfrac{|13|}{\sqrt{25 + 144}}
]
4. Simplify:
[
d = \dfrac{13}{\sqrt{169}} = \dfrac{13}{13}
]
Answer:
[
[d = 1]
]
Question 3: Find the perpendicular distance of [(2, −1)] from the line [x − 2y + 4 = 0].
Step-by-Step Solution:
1. [A = 1, B = −2, C = 4]
2. Point [(2, −1)]
3. Apply formula:
[
d = \dfrac{|1(2) − 2(−1) + 4|}{\sqrt{1^2 + (−2)^2}}
]
4. Simplify numerator:
[
|2 + 2 + 4| = |8| = 8
]
5. Simplify denominator:
[
\sqrt{1 + 4} = \sqrt{5}
]
Answer:
[d = \dfrac{8}{\sqrt{5}}]
Question 4:Â Find the distance of point [(1, 2)] from the line [y = 3x + 1].
Step-by-Step Solution:
1. Convert to general form:
[
3x − y + 1 = 0
]
2. [A = 3, B = −1, C = 1]
3. Apply formula:
[
d = \dfrac{|3(1) − 1(2) + 1|}{\sqrt{9 + 1}}
]
4. Simplify:
[
d = \dfrac{|3 − 2 + 1|}{\sqrt{10}} = \dfrac{2}{\sqrt{10}}
]
Answer:
[d = \dfrac{2}{\sqrt{10}}]
Question 5: Find the perpendicular distance of [(−2, 3)] from [4x + 3y = 12].
Step-by-Step Solution:
1. Convert equation:
[
4x + 3y − 12 = 0
]
2. [A = 4, B = 3, C = −12]
3. Apply formula:
[
d = \dfrac{|4(−2) + 3(3) − 12|}{\sqrt{16 + 9}}
]
4. Simplify:
[
|−8 + 9 − 12| = |−11| = 11
]
5. Denominator:
[
\sqrt{25} = 5
]
Answer:
[d = \dfrac{11}{5}]
Question 6: Find the distance of [(4, −3)] from the line [2x − y − 5 = 0].
Step-by-Step Solution:
1. [A = 2, B = −1, C = −5]
2. Apply formula:
[
d = \dfrac{|2(4) − 1(−3) − 5|}{\sqrt{4 + 1}}
]
3. Simplify:
[
|8 + 3 − 5| = |6| = 6
]
4. Denominator:
[
\sqrt{5}
]
Answer:
[d = \dfrac{6}{\sqrt{5}}]
Question 7: Find the distance of the point [(a, b)] from the line [ax + by + c = 0].
Step-by-Step Solution:
1. Substitute [(x_1, y_1) = (a, b)]
2. Apply formula:
[
d = \dfrac{|a^2 + b^2 + c|}{\sqrt{a^2 + b^2}}
]
**Answer:**
[d = \dfrac{|a^2 + b^2 + c|}{\sqrt{a^2 + b^2}}]
Question 8: Find the value of [k] such that the distance of [(2, 1)] from [kx − y + 5 = 0] is [\sqrt{5}].
Step-by-Step Solution:
1. Apply distance formula:
[
\dfrac{|2k − 1 + 5|}{\sqrt{k^2 + 1}} = \sqrt{5}
]
2. Simplify:
[
\dfrac{|2k + 4|}{\sqrt{k^2 + 1}} = \sqrt{5}
]
3. Square both sides:
[
\dfrac{(2k + 4)^2}{k^2 + 1} = 5
]
4. Cross multiply:
[
(2k + 4)^2 = 5(k^2 + 1)
]
5. Expand:
[
4k^2 + 16k + 16 = 5k^2 + 5
]
6. Rearranging:
[
k^2 − 16k − 11 = 0
]
Answer:
[k = 8 \pm \sqrt{75}]
Question 9: Show that the distance of the origin from [ax + by + c = 0] is independent of the slope of the line.
Step-by-Step Solution:
1. Distance from origin:
[
d = \dfrac{|c|}{\sqrt{a^2 + b^2}}
]
2. Slope of line:
[
m = −\dfrac{a}{b}
]
3. Distance expression does not involve [m].
Conclusion:
Distance depends on position, not inclination.
Question 10: If the distance of [(1, −2)] from the line [3x + 4y + k = 0] is 2, find [k].
Step-by-Step Solution:
1. Apply formula:
[
\dfrac{|3(1) + 4(−2) + k|}{5} = 2
]
2. Simplify:
[
|3 − 8 + k| = 10
]
3. Two cases:
* [k − 5 = 10 ⇒ k = 15]
* [k − 5 = −10 ⇒ k = −5]
Answer:
[k = 15 \text{ or } −5]
Sign in with Google