A straight line is a set of points in a plane that satisfy a linear equation in x and y. Its position can be described through slope, intercepts, inclination, and distance from a point.
A line looks simple, but its equation can change form based on the data given. Important Questions Class 11 Maths Chapter 9 help students practise slope, inclination, angle between lines, parallel and perpendicular conditions, point-slope form, two-point form, slope-intercept form, intercept form, general equation, distance from a point, and distance between parallel lines. The 2026 NCERT chapter Straight Lines connects algebra with coordinate geometry through equations of lines and slope-based reasoning.
Key Takeaways
- Slope Formula: For points (x₁, y₁) and (x₂, y₂), slope is m = (y₂ − y₁)/(x₂ − x₁).
- Parallel Lines: Two non-vertical lines are parallel when their slopes are equal.
- Perpendicular Lines: Two non-vertical lines are perpendicular when m₁m₂ = −1.
- Point-Line Distance: Distance from (x₁, y₁) to Ax + By + C = 0 is |Ax₁ + By₁ + C|/√(A² + B²).
Important Questions Class 11 Maths Chapter 9 Structure 2026
| Concept |
Formula Or Rule |
Key Variables |
| Slope and angle |
m = tan θ; m = (y₂ − y₁)/(x₂ − x₁) |
Inclination θ, slope m |
| Equation of line |
y − y₁ = m(x − x₁); y = mx + c; x/a + y/b = 1 |
Point, slope, intercepts |
| Distance formula for line |
`d = |
Ax₁ + By₁ + C |
Straight Lines Class 11 Chapter Overview
Coordinate geometry studies shapes using algebraic equations. A straight line becomes an equation whose points satisfy one fixed linear condition.
NCERT begins with slope because slope controls direction. It then builds different line equations from points, intercepts, angles, and distances.
Q1. What Is Straight Lines Class 11 About?
Straight Lines Class 11 studies how a line is represented algebraically in the coordinate plane.
The chapter connects slope, inclination, intercepts, and point conditions. It uses equations to check whether a point lies on a line.
Example: y = 2x + 3 represents a straight line.
Q2. Why Are Important Questions Class 11 Maths Chapter 9 Useful For Exams?
Important Questions Class 11 Maths Chapter 9 are useful because the chapter tests formulas and equation selection.
Students must choose the right form of a line from the given data. They also solve distance, angle, parallel, and perpendicular line questions.
Example: Given slope and point, use y − y₁ = m(x − x₁).
Q3. What Is The General Equation Of A Line Class 11?
The general equation of a line is Ax + By + C = 0.
Here, A and B cannot both be zero. This form helps find slope, intercepts, and distance from a point.
Example: 3x − 4y + 7 = 0 is a general linear equation.
Slope Of A Line Class 11 Important Questions
Slope measures the steepness and direction of a line. It comes from the tangent of the line’s inclination.
A horizontal line has slope 0. A vertical line has undefined slope.
Q4. What Is Slope Of A Line Class 11?
Slope of a line is tan θ, where θ is the inclination of the line.
The inclination is measured anticlockwise from the positive x-axis. Slope is denoted by m.
Formula:
m = tan θ, where θ ≠ 90°
Q5. Find The Slope Of A Line Passing Through (3, −2) And (−1, 4).
The slope of the line is −3/2.
- Given Data:
(x₁, y₁) = (3, −2)
(x₂, y₂) = (−1, 4)
- Formula Used:
m = (y₂ − y₁)/(x₂ − x₁)
- Calculation:
m = [4 − (−2)]/[−1 − 3]
m = 6/(−4)
m = −3/2
Final Result: m = −3/2
Q6. Find The Slope Of A Line Passing Through (3, −2) And (7, −2).
The slope of the line is 0.
- Given Data:
(3, −2) and (7, −2)
- Formula Used:
m = (y₂ − y₁)/(x₂ − x₁)
- Calculation:
m = [−2 − (−2)]/(7 − 3)
m = 0/4
m = 0
Final Result: m = 0
Q7. Find The Slope Of A Line Passing Through (3, −2) And (3, 4).
The slope of the line is undefined.
- Given Data:
(3, −2) and (3, 4)
- Formula Used:
m = (y₂ − y₁)/(x₂ − x₁)
- Calculation:
m = [4 − (−2)]/(3 − 3)
m = 6/0
- Since division by zero is not defined, the slope is undefined.
Final Result: Slope is undefined
Q8. Find The Slope Of A Line Making 60° With The Positive x-Axis.
The slope of the line is √3.
- Given Data:
Inclination θ = 60°
- Formula Used:
m = tan θ
- Calculation:
m = tan 60°
m = √3
Final Result: m = √3
Inclination Of A Line Class 11 Questions
Inclination tells the angle a line makes with the positive x-axis. NCERT measures it anticlockwise.
The value of inclination lies from 0° to 180°. A vertical line has inclination 90°.
Q9. What Is Inclination Of A Line Class 11?
Inclination of a line is the angle made with the positive direction of the x-axis.
It is measured anticlockwise. Its value satisfies 0° ≤ θ ≤ 180°.
Example: The x-axis has inclination 0°.
Q10. What Is The Slope Of The x-Axis And y-Axis?
The slope of the x-axis is 0, and the slope of the y-axis is undefined.
The x-axis has inclination 0°. The y-axis has inclination 90°.
Calculation:
tan 0° = 0
tan 90° is not defined.
Final Result: x-axis slope = 0, y-axis slope = undefined
Q11. Find The Slope Of A Line Which Makes 30° With The Positive y-Axis Anticlockwise.
The slope of the line is −√3.
- Given Data:
Line makes 30° with positive y-axis anticlockwise.
- Inclination from positive x-axis:
θ = 90° + 30° = 120°
- Formula Used:
m = tan θ
- Calculation:
m = tan 120°
m = −√3
Final Result: m = −√3
Parallel And Perpendicular Lines Class 11 Important Questions
Slopes decide whether two non-vertical lines are parallel or perpendicular. This avoids drawing diagrams.
Equal slopes give parallel lines. Negative reciprocal slopes give perpendicular lines.
Q12. What Is The Condition For Parallel Lines Class 11?
Two non-vertical lines are parallel if and only if their slopes are equal.
If line slopes are m₁ and m₂, then parallel lines satisfy m₁ = m₂.
Example: Lines with slopes 2 and 2 are parallel.
Q13. What Is The Condition For Perpendicular Lines Class 11?
Two non-vertical lines are perpendicular if and only if m₁m₂ = −1.
This means one slope is the negative reciprocal of the other. The angle between the lines is 90°.
Example: Slopes 3 and −1/3 are perpendicular.
Q14. Line Through (−2, 6) And (4, 8) Is Perpendicular To Line Through (8, 12) And (x, 24). Find x.
The value of x is 4.
- Slope of first line:
m₁ = (8 − 6)/(4 − (−2))
m₁ = 2/6 = 1/3
- Slope of second line:
m₂ = (24 − 12)/(x − 8)
m₂ = 12/(x − 8)
- Perpendicular condition:
m₁m₂ = −1
- Substitute:
(1/3) × [12/(x − 8)] = −1
- Solve:
4/(x − 8) = −1
x − 8 = −4
x = 4
Final Result: x = 4
Q15. Show That Points (−2, −1), (4, 0), (3, 3), And (−3, 2) Form A Parallelogram Without Distance Formula.
The points form a parallelogram because opposite sides have equal slopes.
- Let:
A(−2, −1), B(4, 0), C(3, 3), D(−3, 2)
- Slope of AB:
mAB = [0 − (−1)]/[4 − (−2)] = 1/6
- Slope of CD:
mCD = (2 − 3)/(−3 − 3) = −1/−6 = 1/6
- Slope of BC:
mBC = (3 − 0)/(3 − 4) = −3
- Slope of AD:
mAD = [2 − (−1)]/[−3 − (−2)] = 3/−1 = −3
Final Result: AB ∥ CD and BC ∥ AD
Angle Between Two Lines Class 11 Questions
The angle between two lines comes from their slopes. NCERT gives the tangent formula for the acute angle.
Use absolute value when the question asks for the acute angle.
Q16. What Is The Formula For Angle Between Two Lines Class 11?
The acute angle θ between two lines with slopes m₁ and m₂ is given by tan θ = |(m₂ − m₁)/(1 + m₁m₂)|.
The formula works when 1 + m₁m₂ ≠ 0. If 1 + m₁m₂ = 0, lines are perpendicular.
Formula:
tan θ = |(m₂ − m₁)/(1 + m₁m₂)|
Q17. If The Angle Between Two Lines Is 45° And One Slope Is 1/2, Find The Other Slope.
The other slope is 3 or −1/3.
- Given Data:
θ = 45°
m₁ = 1/2
Let m₂ = m
- Formula Used:
tan θ = |(m − 1/2)/(1 + m/2)|
- Since tan 45° = 1:
|(m − 1/2)/(1 + m/2)| = 1
- Case 1:
(m − 1/2)/(1 + m/2) = 1
m − 1/2 = 1 + m/2
m = 3
- Case 2:
(m − 1/2)/(1 + m/2) = −1
m − 1/2 = −1 − m/2
m = −1/3
Final Result: m = 3 or m = −1/3
Q18. Find The Angle Between x-Axis And Line Joining (3, −1) And (4, −2).
The angle is 135°.
- Given Points:
(3, −1) and (4, −2)
- Find slope:
m = [−2 − (−1)]/(4 − 3)
m = −1/1 = −1
- Use m = tan θ:
tan θ = −1
- Since 0° ≤ θ ≤ 180°:
θ = 135°
Final Result: 135°
Point Slope Form Class 11 Questions
Point-slope form works when one point and the slope are given. It is often the shortest form.
Every point on the line must satisfy y − y₁ = m(x − x₁).
Q19. What Is Point Slope Form Class 11?
Point-slope form is y − y₁ = m(x − x₁).
It gives the equation of a non-vertical line passing through (x₁, y₁) with slope m.
Example: Through (−2, 3) with slope −4, use y − 3 = −4(x + 2).
Q20. Find The Equation Of A Line Through (−2, 3) With Slope −4.
The equation is 4x + y + 5 = 0.
- Given Data:
Point = (−2, 3)
Slope = −4
- Formula Used:
y − y₁ = m(x − x₁)
- Substitute values:
y − 3 = −4[x − (−2)]
- Simplify:
y − 3 = −4(x + 2)
y − 3 = −4x − 8
4x + y + 5 = 0
Final Result: 4x + y + 5 = 0
Q21. A Line Passes Through (x₁, y₁) And (h, k). If Its Slope Is m, Show k − y₁ = m(h − x₁).
The required relation follows directly from point-slope form.
- Given Data:
Point on line = (x₁, y₁)
Another point = (h, k)
Slope = m
- Point-slope form:
y − y₁ = m(x − x₁)
- Substitute (h, k):
k − y₁ = m(h − x₁)
Final Result: k − y₁ = m(h − x₁)
Two Point Form Class 11 Important Questions
Two-point form works when two points on the line are known. It avoids first finding intercepts.
The slope comes from the same two points.
Q22. What Is Two Point Form Class 11?
Two-point form gives the equation of a line passing through (x₁, y₁) and (x₂, y₂).
The formula uses the slope between the two points. It works when x₁ ≠ x₂.
Formula:
y − y₁ = [(y₂ − y₁)/(x₂ − x₁)](x − x₁)
Q23. Find The Equation Of The Line Through (1, −1) And (3, 5).
The equation is −3x + y + 4 = 0.
- Given Data:
(x₁, y₁) = (1, −1)
(x₂, y₂) = (3, 5)
- Formula Used:
y − y₁ = [(y₂ − y₁)/(x₂ − x₁)](x − x₁)
- Substitute values:
y − (−1) = [5 − (−1)]/(3 − 1) × (x − 1)
- Calculate:
y + 1 = 6/2 × (x − 1)
y + 1 = 3x − 3
- Rearrange:
−3x + y + 4 = 0
Final Result: −3x + y + 4 = 0
Q24. Find The Equation Of The Median Through R For Triangle P(2, 1), Q(−2, 3), R(4, 5).
The median through R has equation x + 3y − 19 = 0.
- Find midpoint of PQ:
M = ((2 + (−2))/2, (1 + 3)/2)
M = (0, 2)
- Median passes through:
R(4, 5) and M(0, 2)
- Slope of RM:
m = (5 − 2)/(4 − 0) = 3/4
- Use point-slope form through (0, 2):
y − 2 = (3/4)(x − 0)
- Rearrange:
4y − 8 = 3x
3x − 4y + 8 = 0
Final Result: 3x − 4y + 8 = 0
Slope Intercept Form Class 11 Questions
Slope-intercept form gives a line from its slope and y-intercept. It is useful when the line cuts the y-axis at a known point.
The constant c is the y-intercept in y = mx + c.
Q25. What Is Slope Intercept Form Class 11?
Slope-intercept form is y = mx + c.
Here, m is the slope and c is the y-intercept. The line meets the y-axis at (0, c).
Example: y = 2x − 3 has slope 2 and y-intercept −3.
Q26. Find The Equation When tan θ = 1/2 And y-Intercept Is −3/2.
The equation is x − 2y − 3 = 0.
- Given Data:
m = tan θ = 1/2
c = −3/2
- Formula Used:
y = mx + c
- Substitute values:
y = (1/2)x − 3/2
- Multiply by 2:
2y = x − 3
- Rearrange:
x − 2y − 3 = 0
Final Result: x − 2y − 3 = 0
Q27. Find The Equation Of A Line With Slope 1/2 And x-Intercept 4.
The equation is x − 2y − 4 = 0.
- Given Data:
Slope m = 1/2
x-intercept d = 4
- Formula Used:
y = m(x − d)
- Substitute values:
y = (1/2)(x − 4)
- Multiply by 2:
2y = x − 4
- Rearrange:
x − 2y − 4 = 0
Final Result: x − 2y − 4 = 0
Intercept Form Of A Line Class 11 Questions
Intercept form works when x-intercept and y-intercept are given. It uses the points where a line cuts the axes.
The formula is x/a + y/b = 1.
Q28. What Is Intercept Form Of A Line Class 11?
Intercept form of a line is x/a + y/b = 1.
Here, a is the x-intercept and b is the y-intercept. Both intercepts must be non-zero.
Example: If intercepts are −3 and 2, use x/(−3) + y/2 = 1.
Q29. Find The Equation Of The Line With x-Intercept −3 And y-Intercept 2.
The equation is 2x − 3y + 6 = 0.
- Given Data:
a = −3
b = 2
- Formula Used:
x/a + y/b = 1
- Substitute:
x/(−3) + y/2 = 1
- Multiply by 6:
−2x + 3y = 6
- Rearrange:
2x − 3y + 6 = 0
Final Result: 2x − 3y + 6 = 0
Q30. Find The Equation Of A Line That Cuts Equal Intercepts And Passes Through (2, 3).
The equation is x + y = 5.
- Equal intercepts mean:
a = b
- Intercept form becomes:
x/a + y/a = 1
- Simplify:
x + y = a
- Since point (2, 3) lies on the line:
2 + 3 = a
a = 5
Final Result: x + y = 5
General Equation Of A Line Class 11 Questions
The general form Ax + By + C = 0 is useful for converting between forms. It also supports distance formula.
From this form, slope is −A/B when B ≠ 0.
Q31. How Do You Find Slope From General Equation Of A Line Class 11?
For Ax + By + C = 0, slope is −A/B when B ≠ 0.
Move the equation into y = mx + c form. The coefficient of x becomes the slope.
Example: 3x − 4y + 7 = 0 gives slope 3/4.
Q32. Reduce 6x + 3y − 5 = 0 To Slope-Intercept Form.
The slope-intercept form is y = −2x + 5/3.
- Given equation:
6x + 3y − 5 = 0
- Move terms:
3y = −6x + 5
- Divide by 3:
y = −2x + 5/3
- Compare with y = mx + c:
m = −2
c = 5/3
Final Result: slope = −2, y-intercept = 5/3
Q33. Reduce 3x + 2y − 12 = 0 To Intercept Form.
The intercept form is x/4 + y/6 = 1.
- Given equation:
3x + 2y − 12 = 0
- Move constant:
3x + 2y = 12
- Divide by 12:
x/4 + y/6 = 1
- Identify intercepts:
x-intercept = 4
y-intercept = 6
Final Result: x/4 + y/6 = 1
Distance Of A Point From A Line Class 11 Important Questions
The distance from a point to a line means the perpendicular distance. It is always non-negative.
NCERT derives this formula using triangle area and intercepts.
Q34. What Is Distance Of A Point From A Line Class 11?
The distance from (x₁, y₁) to Ax + By + C = 0 is |Ax₁ + By₁ + C|/√(A² + B²).
The numerator uses the point in the line equation. The denominator normalises the result.
Formula:
d = |Ax₁ + By₁ + C|/√(A² + B²)
Q35. Find The Distance Of (3, −5) From The Line 3x − 4y − 26 = 0.
The distance is 3 units.
- Given Data:
Point = (3, −5)
Line = 3x − 4y − 26 = 0
- Identify values:
A = 3, B = −4, C = −26
- Formula Used:
d = |Ax₁ + By₁ + C|/√(A² + B²)
- Substitute:
d = |3(3) + (−4)(−5) − 26|/√(3² + (−4)²)
- Calculate:
d = |9 + 20 − 26|/√25
d = 3/5 × 5
d = 3
Final Result: 3 units
Q36. Find The Distance Of (−1, 1) From 12(x + 6) = 5(y − 2).
The distance is 5 units.
- Convert line to general form:
12x + 72 = 5y − 10
12x − 5y + 82 = 0
- Point:
(−1, 1)
- Formula Used:
d = |Ax₁ + By₁ + C|/√(A² + B²)
- Substitute:
d = |12(−1) − 5(1) + 82|/√(12² + (−5)²)
- Calculate:
d = |−12 − 5 + 82|/√169
d = 65/13 = 5
Final Result: 5 units
Distance Between Two Parallel Lines Class 11 Questions
Parallel lines keep a constant perpendicular distance. Their general form must have the same A and B coefficients.
If lines are Ax + By + C₁ = 0 and Ax + By + C₂ = 0, use only constants in the numerator.
Q37. What Is The Formula For Distance Between Two Parallel Lines Class 11?
The distance between Ax + By + C₁ = 0 and Ax + By + C₂ = 0 is |C₁ − C₂|/√(A² + B²).
The formula applies only when A and B match in both lines. If needed, first make coefficients identical.
Formula:
d = |C₁ − C₂|/√(A² + B²)
Q38. Find The Distance Between 3x − 4y + 7 = 0 And 3x − 4y + 5 = 0.
The distance between the lines is 2/5 units.
- Given Lines:
3x − 4y + 7 = 0
3x − 4y + 5 = 0
- Identify values:
A = 3, B = −4, C₁ = 7, C₂ = 5
- Formula Used:
d = |C₁ − C₂|/√(A² + B²)
- Substitute:
d = |7 − 5|/√(3² + (−4)²)
- Calculate:
d = 2/√25 = 2/5
Final Result: 2/5 units
Q39. Find The Distance Between 15x + 8y − 34 = 0 And 15x + 8y + 31 = 0.
The distance between the lines is 65/17 units.
- Given Lines:
15x + 8y − 34 = 0
15x + 8y + 31 = 0
- Identify values:
A = 15, B = 8, C₁ = −34, C₂ = 31
- Formula Used:
d = |C₁ − C₂|/√(A² + B²)
- Substitute:
d = |−34 − 31|/√(15² + 8²)
- Calculate:
d = 65/√289
d = 65/17
Final Result: 65/17 units
NCERT Class 11 Maths Chapter 9 Questions On Applications
NCERT application questions use equations of lines in real situations. The chapter includes temperature, milk demand, image reflection, and shortest path cases.
These questions test whether students can translate a condition into a line equation.
Q40. A Copper Rod Has L = 124.942 At C = 20 And L = 125.134 At C = 110. Express L In Terms Of C.
The linear relation is L = 0.002133C + 124.8993 approximately.
- Given Points:
(C, L) = (20, 124.942)
(110, 125.134)
- Find slope:
m = (125.134 − 124.942)/(110 − 20)
m = 0.192/90
m = 0.002133
- Use point-slope form:
L − 124.942 = 0.002133(C − 20)
- Simplify:
L = 0.002133C + 124.8993
Final Result: L = 0.002133C + 124.8993
Q41. A Milk Store Sells 980 Litres At ₹14 Per Litre And 1220 Litres At ₹16 Per Litre. Estimate Demand At ₹17.
The store can sell 1340 litres at ₹17 per litre.
- Take price as x and demand as y.
Points are (14, 980) and (16, 1220).
- Find slope:
m = (1220 − 980)/(16 − 14)
m = 240/2 = 120
- Use point-slope form:
y − 980 = 120(x − 14)
- Put x = 17:
y − 980 = 120(3)
y = 1340
Final Result: 1340 litres
Q42. Find The Image Of (1, 2) In The Line x − 3y + 4 = 0.
The image is (6/5, 7/5).
- Let image be:
Q(h, k)
- Mirror line is perpendicular bisector of PQ.
Slope of mirror line x − 3y + 4 = 0 is 1/3.
- Slope of PQ:
−3
- Equation through (1, 2):
k − 2 = −3(h − 1)
3h + k = 5
- Midpoint of P and Q lies on mirror line:
((1+h)/2) − 3((2+k)/2) + 4 = 0
- Simplify with 3h + k = 5:
h = 6/5, k = 7/5
Final Result: Image = (6/5, 7/5)
Q43. If Lines 2x + y − 3 = 0, 5x + ky − 3 = 0, And 3x − y − 2 = 0 Are Concurrent, Find k.
The value of k is −2.
- Solve first and third lines:
2x + y − 3 = 0
3x − y − 2 = 0
- Add equations:
5x − 5 = 0
x = 1
- Substitute in first line:
2(1) + y − 3 = 0
y = 1
- Point of intersection is (1, 1).
- Substitute in second line:
5(1) + k(1) − 3 = 0
k + 2 = 0
k = −2
Final Result: k = −2
Straight Lines Class 11 Board Pattern Questions
Board pattern questions often mix slope, intercepts, and distance formula. The fastest method starts by identifying the given condition.
Use slope for direction, intercept form for axis cuts, and distance formula for perpendicular length.
Q44. Find The Equation Of A Line Passing Through (−3, 5) And Perpendicular To The Line Through (2, 5) And (−3, 6).
The required equation is 5y − x − 28 = 0.
- Slope of given line:
m = (6 − 5)/(−3 − 2)
m = 1/(−5) = −1/5
- Slope of perpendicular line:
m₂ = 5
- Use point-slope form through (−3, 5):
y − 5 = 5(x + 3)
- Simplify:
y − 5 = 5x + 15
y = 5x + 20
- General form:
5x − y + 20 = 0
Final Result: 5x − y + 20 = 0
Q45. Prove That Points (3, 0), (−2, −2), And (8, 2) Are Collinear Using Equation Of A Line.
The points are collinear because all three satisfy the same line equation.
- Find line through (3, 0) and (−2, −2).
- Slope:
m = [−2 − 0]/[−2 − 3]
m = −2/−5 = 2/5
- Use point-slope form:
y − 0 = (2/5)(x − 3)
- Rearrange:
5y = 2x − 6
2x − 5y − 6 = 0
- Test (8, 2):
2(8) − 5(2) − 6 = 16 − 10 − 6 = 0
Final Result: The three points are collinear
Important Questions Class 11 Maths Chapter-Wise