Physics and Mathematics
Example 1 – Distance Between Two Parallel Straight Lines
Practice Questions with Step-by-Step Solutions
Question 1: Find the distance between the parallel lines [4x – 3y + 1 = 0] and [4x – 3y – 11 = 0].
Step-by-Step Solution:
1. Compare with general form [ax + by + c = 0]
[a = 4,][ b = -3,][ c_1 = 1,][ c_2 = -11]
2. Distance formula:
[d = \dfrac{|c_1 – c_2|}{\sqrt{a^2 + b^2}}]
3. Substitute values:
[d = \dfrac{|1 – (-11)|}{\sqrt{4^2 + (-3)^2}}]
4. Simplify:
[d = \dfrac{12}{\sqrt{16 + 9}}]
[d = \dfrac{12}{5}]
Answer:
[d = \dfrac{12}{5}]
Question 2: Find the distance between the lines [7x + y – 4 = 0] and [7x + y + 10 = 0].
Step-by-Step Solution:
1. Identify coefficients:
[a = 7,][ b = 1,][ c_1 = -4,][ c_2 = 10]
2. Apply formula:
[d = \dfrac{|-4 – 10|}{\sqrt{49 + 1}}]
3. Simplify:
[d = \dfrac{14}{\sqrt{50}}]
Answer:
[d = \dfrac{14}{\sqrt{50}}]
Question 3: Find the distance between the parallel lines [x – 3y + 6 = 0] and [x – 3y – 9 = 0].
Step-by-Step Solution:
1. [a = 1,][ b = -3,][ c_1 = 6,][ c_2 = -9]
2. Distance formula:
[d = \dfrac{|6 – (-9)|}{\sqrt{1 + 9}}]
3. Simplify:
[d = \dfrac{15}{\sqrt{10}}]
Answer:
[d = \dfrac{15}{\sqrt{10}}]
Question 4: Find the distance between the lines [y = 2x + 5] and [y = 2x – 3].
Step-by-Step Solution:
1. Convert to general form:
[2x – y + 5 = 0] and [2x – y – 3 = 0]
2. [a = 2,][ b = -1,][ c_1 = 5,][ c_2 = -3]
3. Apply formula:
[d = \dfrac{|5 – (-3)|}{\sqrt{4 + 1}}]
4. Simplify:
[d = \dfrac{8}{\sqrt{5}}]
Answer:
[d = \dfrac{8}{\sqrt{5}}]
Question 5: Find the distance between the lines [3x + 5y = 15] and [3x + 5y = -10].
Step-by-Step Solution:
1. Convert to general form:
[3x + 5y – 15 = 0] and [3x + 5y + 10 = 0]
2. [a = 3,][ b = 5,][ c_1 = -15,][ c_2 = 10]
3. Apply formula:
[d = \dfrac{|-15 – 10|}{\sqrt{9 + 25}}]
4. Simplify:
[d = \dfrac{25}{\sqrt{34}}]
Answer:
[d = \dfrac{25}{\sqrt{34}}]
Question 6: Find the distance between the lines [6x – 8y + 3 = 0] and [3x – 4y – 5 = 0].
Step-by-Step Solution:
1. Normalize second equation by multiplying by 2:
[6x – 8y – 10 = 0]
2. Now:
[c_1 = 3,][ c_2 = -10]
3. Apply formula:
[d = \dfrac{|3 – (-10)|}{\sqrt{36 + 64}}]
4. Simplify:
[d = \dfrac{13}{\sqrt{100}}]
[d = \dfrac{13}{10}]
Answer:
[d = \dfrac{13}{10}]
Question 7: Find the distance between the lines [y = mx + 4] and [y = mx – 6].
Step-by-Step Solution:
1. Convert to general form:
[mx – y + 4 = 0] and [mx – y – 6 = 0]
2. Apply formula:
[d = \dfrac{|4 – (-6)|}{\sqrt{m^2 + 1}}]
Answer:
[d = \dfrac{10}{\sqrt{m^2 + 1}}]
Question 8: Find the distance between the parallel lines [ax + by + c = 0] and [ax + by + d = 0].
Step-by-Step Solution:
1. Coefficients of x and y are identical ⇒ lines are parallel
2. Apply formula:
[d = \dfrac{|c – d|}{\sqrt{a^2 + b^2}}]
Answer:
[d = \dfrac{|c – d|}{\sqrt{a^2 + b^2}}]
Question 9: If the distance between the lines [5x – 12y + 9 = 0] and [5x – 12y + k = 0] is 3, find k.
Step-by-Step Solution:
1. Apply formula:
[\dfrac{|9 – k|}{\sqrt{25 + 144}} = 3]
2. Simplify:
[\dfrac{|9 – k|}{13} = 3]
3. Multiply both sides:
[|9 – k| = 39]
4. Two cases:
[9 – k = 39 ⇒ k = -30]
[9 – k = -39 ⇒ k = 48]
Answer:
[k = -30 or 48]
Question 10: Find the distance between the parallel lines [x = 4] and [x = -7].
Step-by-Step Solution:
1. Distance between vertical lines [x = c_1] and [x = c_2] is:
[d = |c_1 – c_2|]
2. Substitute:
[d = |4 – (-7)|]
Answer:
[d = 11]
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