This MCQ module is based on: Distance from a Line, Distance Between Parallel Lines and Exercises
TOPIC 29 OF 45
Distance from a Line, Distance Between Parallel Lines and Exercises
🎓 Class 11
Mathematics
CBSE
Theory
Ch 9 — Straight Lines
⏱ ~30 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: Distance from a Line, Distance Between Parallel Lines and Exercises
Targeting Class 11 level in Coordinate Geometry, with Advanced difficulty.
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9.5 Distance of a Point from a Line
The perpendicular distancei of a point \(P(x_1,y_1)\) from the line \(L:Ax+By+C=0\) is the shortest (perpendicular) distance \(d\) from \(P\) to \(L\).
Derivation
Let \(L\) cut the \(x\)-axis at \(Q(-C/A,0)\) and \(y\)-axis at \(R(0,-C/B)\). Drop perpendicular \(PM\) from \(P\) onto \(L\). Using area of triangle \(PQR\) in two ways (base–height vs. the shoelace determinant):
\(\text{Area}=\tfrac{1}{2}\cdot QR\cdot PM=\tfrac12 \left|x_1\left(0-(-C/B)\right)+(-C/A)\left(-C/B-y_1\right)+0\cdot(y_1-0)\right|.\)
Simplify and note \(QR=\sqrt{(C/A)^2+(C/B)^2}=\dfrac{|C|\sqrt{A^2+B^2}}{|AB|}\). After cancellation:
Formula
\[d=\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}.\] Set numerator to 0 \(\iff\) the point lies on the line.Fig 9.8 — Perpendicular distance \(d\) from \(P\) to line \(L\).
9.6 Distance Between Two Parallel Lines
Two parallel lines have the same slope. Writing them as \(Ax+By+C_1=0\) and \(Ax+By+C_2=0\), the distance between them equals the perpendicular distance from any point on the first line to the second:
Formula
\[d=\frac{|C_1-C_2|}{\sqrt{A^2+B^2}}.\]Fig 9.9 — Distance between two parallel lines with same \(A,B\).
Example 13
Find the distance of the point \((3,-5)\) from the line \(3x-4y-26=0\).\(d=\dfrac{|3(3)-4(-5)-26|}{\sqrt{9+16}}=\dfrac{|9+20-26|}{5}=\dfrac{3}{5}.\)
Example 14
Find the distance between the parallel lines \(3x-4y+7=0\) and \(3x-4y+5=0\).\(d=\dfrac{|7-5|}{\sqrt{9+16}}=\dfrac{2}{5}.\)
Example 15
Find the distance between \(4x+3y=11\) and \(8x+6y=15\).Rewrite the second as \(4x+3y=\tfrac{15}{2}\). Here \(C_1=-11,C_2=-\tfrac{15}{2}\). \(d=\dfrac{|-11+15/2|}{\sqrt{16+9}}=\dfrac{|{-7/2}|}{5}=\dfrac{7}{10}.\)
Activity 9.3 — Highway buffer zone
Predict: Two railway tracks are modelled by \(3x+4y-7=0\) and \(3x+4y-17=0\). Before calculating, estimate the buffer width between them.
- Compute \(\sqrt{A^2+B^2}=\sqrt{9+16}=5\).
- Find the perpendicular distance using \(d=|C_1-C_2|/5\).
- If a safety regulation requires the distance to be at least \(2.5\) units, does the current layout comply?
- Rewrite the second line so the buffer becomes exactly \(3\) units.
Insight: \(d=|{-7}-({-17})|/5=10/5=2\). Not compliant with a 2.5 unit rule. To make \(d=3\), we need \(|C_1-C_2|=15\), so change the second equation to \(3x+4y-22=0\).
Exercise 9.1 — End-of-Chapter Selected Questions
Q1. Find the distance of \((-1,1)\) from the line \(12(x+6)=5(y-2)\).
Expand: \(12x+72=5y-10\Rightarrow 12x-5y+82=0\). \(d=\dfrac{|12(-1)-5(1)+82|}{\sqrt{144+25}}=\dfrac{|65|}{13}=5.\)
Q2. Find the points on the \(x\)-axis whose distances from the line \(\tfrac{x}{3}+\tfrac{y}{4}=1\) are 4 units.
Rewrite: \(4x+3y-12=0\). Let point be \((a,0)\). \(\dfrac{|4a-12|}{5}=4\Rightarrow |4a-12|=20\Rightarrow 4a-12=\pm20\Rightarrow a=8\) or \(a=-2\). Points: \((8,0),(-2,0)\).
Q3. Find the distance between the parallel lines \(15x+8y-34=0\) and \(15x+8y+31=0\).
\(\sqrt{15^2+8^2}=17\); \(d=\dfrac{|{-34}-31|}{17}=\dfrac{65}{17}.\)
Q4. Find the equation of a line that is parallel to \(3x-4y+2=0\) and passes through \((-2,3)\).
Same slope: \(3x-4y+k=0\). Through \((-2,3)\): \(-6-12+k=0\Rightarrow k=18\). Line: \(3x-4y+18=0.\)
Q5. Find the equation of a line perpendicular to \(x-2y+3=0\) and passing through \((1,-2)\).
Given slope \(=\tfrac12\); perpendicular slope \(=-2\). Line: \(y+2=-2(x-1)\Rightarrow 2x+y=0.\)
Q6. Find the angle between the lines \(y-\sqrt3 x-5=0\) and \(\sqrt3 y-x+6=0\).
Slopes: \(m_1=\sqrt3,m_2=1/\sqrt3\). \(\tan\theta=\left|\dfrac{1/\sqrt3-\sqrt3}{1+\sqrt3\cdot 1/\sqrt3}\right|=\left|\dfrac{-2/\sqrt3}{2}\right|=\dfrac{1}{\sqrt3}\Rightarrow \theta=30°.\)
Q7. If the lines \(2x+y-3=0,\ 5x+ky-3=0\) and \(3x-y-2=0\) are concurrent, find \(k\).
Intersection of first and third: solve \(2x+y=3, 3x-y=2\Rightarrow 5x=5\Rightarrow x=1,y=1\). Substitute into second: \(5+k-3=0\Rightarrow k=-2.\)
Q8. Find the equation of the line passing through the point of intersection of \(3x+y-9=0\) and \(4x-y+2=0\), perpendicular to the line \(x+2y=0\).
Intersection: add → \(7x-7=0\Rightarrow x=1\); then \(y=6\). Point \((1,6)\). Perpendicular slope to \(x+2y=0\) (slope \(-1/2\)) is \(2\). Line: \(y-6=2(x-1)\Rightarrow 2x-y+4=0.\)
Q9. Find the value of \(p\) so that the three lines \(3x+y-2=0,\ px+2y-3=0,\ 2x-y-3=0\) may intersect at one point.
Intersection of first and third: add \(\Rightarrow 5x-5=0\Rightarrow x=1,y=-1\). Point \((1,-1)\) lies on second: \(p-2-3=0\Rightarrow p=5.\)
Q10. Find the length of the perpendicular from the origin to the line \(\tfrac{x}{5}+\tfrac{y}{12}=1\).
Rewrite: \(12x+5y-60=0\). \(d=\dfrac{|0+0-60|}{\sqrt{144+25}}=\dfrac{60}{13}.\)
Summary
- Slope \(m=\tan\theta\); two-point formula \(m=\tfrac{y_2-y_1}{x_2-x_1}\).
- Parallel lines: \(m_1=m_2\); perpendicular: \(m_1 m_2=-1\).
- Acute angle between lines: \(\tan\theta=\left|\tfrac{m_2-m_1}{1+m_1m_2}\right|\).
- Point-slope: \(y-y_0=m(x-x_0)\); slope-intercept: \(y=mx+c\); intercept: \(\tfrac{x}{a}+\tfrac{y}{b}=1\); normal: \(x\cos\omega+y\sin\omega=p\).
- General: \(Ax+By+C=0\).
- Distance of \((x_1,y_1)\) from \(Ax+By+C=0\) is \(\tfrac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}\).
- Distance between parallel lines \(Ax+By+C_1=0\) and \(Ax+By+C_2=0\) is \(\tfrac{|C_1-C_2|}{\sqrt{A^2+B^2}}\).
Competency-Based Questions — Surveying and Safety Distances
A municipal authority is laying two parallel pipelines modelled by the lines \(L_1: 5x-12y+26=0\) and \(L_2:5x-12y-13=0\) (coordinates in metres). A residential pillar stands at \(P(4,3)\).
Q1. Compute the perpendicular distance of the pillar \(P(4,3)\) from pipeline \(L_1\).
L3 Apply\(d=\dfrac{|5(4)-12(3)+26|}{\sqrt{25+144}}=\dfrac{|20-36+26|}{13}=\dfrac{10}{13}\) m.
Q2. Find the distance between the two pipelines and interpret whether \(P\) lies within the strip between them.
L4 Analyse\(d=\dfrac{|26-(-13)|}{13}=\dfrac{39}{13}=3\) m. Check signs of \(5x-12y+26\) at \(P\): \(10>0\) (same side as a test origin which gives \(26>0\)). At \(L_2:5x-12y-13=20-36-13=-29<0\). Different signs w.r.t. \(L_2\) strip boundary — \(P\) lies between the two lines.
Q3. A safety guideline demands that no construction stand closer than 0.5 m to either pipeline. Evaluate whether the pillar complies.
L5 EvaluateDistance to \(L_1=\tfrac{10}{13}\approx0.77\) m. Distance to \(L_2=\tfrac{|20-36-13|}{13}=\tfrac{29}{13}\approx2.23\) m. Both exceed 0.5 m — the pillar complies.
Q4. Design a new line parallel to \(L_1\) such that \(P(4,3)\) lies exactly on it, and express it in the form \(5x-12y+C=0\). Verify your answer.
L6 CreateParallel \(\Rightarrow 5x-12y+C=0\). Through \((4,3)\): \(20-36+C=0\Rightarrow C=16\). Line: \(5x-12y+16=0\). Verify: \(5(4)-12(3)+16=0\) ✓.
Assertion–Reason Questions
Assertion (A): The distance of the origin from the line \(3x+4y+5=0\) is 1.
Reason (R): For the line \(Ax+By+C=0\), the distance from origin is \(\tfrac{|C|}{\sqrt{A^2+B^2}}\).
Reason (R): For the line \(Ax+By+C=0\), the distance from origin is \(\tfrac{|C|}{\sqrt{A^2+B^2}}\).
Answer: A. \(d=\tfrac{|5|}{\sqrt{9+16}}=\tfrac{5}{5}=1\); R supplies the formula.
Assertion (A): The distance between the lines \(3x-4y+7=0\) and \(6x-8y-5=0\) is \(\tfrac{19}{10}\).
Reason (R): For distance between two parallel lines, the coefficients of \(x\) and \(y\) must be made equal before applying \(\tfrac{|C_1-C_2|}{\sqrt{A^2+B^2}}\).
Reason (R): For distance between two parallel lines, the coefficients of \(x\) and \(y\) must be made equal before applying \(\tfrac{|C_1-C_2|}{\sqrt{A^2+B^2}}\).
Answer: A. Rewrite the second as \(3x-4y-\tfrac{5}{2}=0\); \(d=\tfrac{|7+5/2|}{5}=\tfrac{19/2}{5}=\tfrac{19}{10}\). R explains the procedural requirement.
Frequently Asked Questions
How do you find the distance between two parallel lines?
Write the parallel lines in the form A x + B y + C1 = 0 and A x + B y + C2 = 0 (same A and B). Distance = |C1 - C2| / root(A squared + B squared).
What is the summary of Class 11 Chapter 9 Straight Lines?
Chapter 9 covers slope, angle between lines, various forms of the equation of a line, distance of a point from a line and distance between parallel lines in the Cartesian plane.
What is the foot of the perpendicular from a point to a line?
It is the point on the line closest to the given external point. The segment joining them is perpendicular to the line.
What is the shift-of-origin technique for lines?
If the origin is shifted to (h, k), the new coordinates (X, Y) satisfy x = X + h, y = Y + k. This can simplify line equations.
Are conic sections in Chapter 10 related to lines?
Yes. Tangents to conic sections and asymptotes of hyperbolas are straight lines, so fluency with Chapter 9 forms is essential for Chapter 10 tangent problems.
How much weightage does Chapter 9 carry in JEE?
Straight lines typically contributes 1-2 questions (4-8 marks) in JEE Main and is a foundation for conic sections and 2D coordinate geometry in JEE Advanced.
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