JEE Main Mathematics · Chapter 12

Straight Lines

Slope, forms of a line, distance of a point from a line, angle between lines, family of lines.

NCERT Class 11JEE Main: 1 Q per shift 9 min read

1. Chapter at a Glance

Coordinate geometry turns geometric shapes into equations. The straight line is the simplest curve — an equation of the first degree, ax + by + c = 0. It is also the foundation for circles and conics, and it appears inside 3D geometry and calculus problems. JEE Main usually has one direct question on lines, plus several questions in other chapters that need these formulas.

What JEE asks most
  • Distance of a point from a line and between parallel lines.
  • Foot of the perpendicular and image of a point in a line.
  • Angle between lines, concurrency, and the family of lines through an intersection.
  • Locus problems and triangle centres (centroid, orthocentre, circumcentre).

2. Basic Formulas of Coordinate Geometry

  • Distance between (x1, y1) and (x2, y2): √((x2 − x1)² + (y2 − y1)²).
  • Section formula (internal, ratio m : n): ((mx2 + nx1)/(m + n), (my2 + ny1)/(m + n)). For external division, replace n by −n.
  • Centroid of a triangle: ((x1 + x2 + x3)/3, (y1 + y2 + y3)/3).
  • Incentre: ((ax1 + bx2 + cx3)/(a + b + c), …), where a, b, c are the side lengths opposite the vertices.
  • Area of a triangle: ½|x1(y2 − y3) + x2(y3 − y1) + x3(y1 − y2)|.
Triangle centres

The centroid G lies on the line joining the orthocentre H and the circumcentre O (the Euler line) and divides HO in the ratio 2 : 1. In a right-angled triangle the orthocentre is the right-angle vertex and the circumcentre is the midpoint of the hypotenuse. In an equilateral triangle all centres coincide.

3. Slope and Forms of a Line

The slope m = tan θ, where θ is the angle the line makes with the positive x-axis. Through two points, m = (y2 − y1)/(x2 − x1).

FormEquationWhen to use
Slope-intercepty = mx + cslope and y-intercept known
Point-slopey − y1 = m(x − x1)a point and slope known
Two-point(y − y1)/(y2 − y1) = (x − x1)/(x2 − x1)two points known
Interceptx/a + y/b = 1intercepts on the axes known
Normalx cos α + y sin α = pperpendicular distance p from origin and its direction α
Parametricx = x1 + r cos θ, y = y1 + r sin θpoints at distance r along the line
Generalax + by + c = 0slope = −a/b
(0, b)(a, 0) p O θ x/a + y/b = 1
Intercept form x/a + y/b = 1; p is the length of the perpendicular from O to the line (normal form).

4. Angle, Distance and Position

Angle between two lines

tan θ = |(m1 − m2)/(1 + m1m2)|

Parallel: m1 = m2. Perpendicular: m1m2 = −1 (or a1a2 + b1b2 = 0).

Distances

Distance of (x1, y1) from ax + by + c = 0 = |ax1 + by1 + c|/√(a² + b²)
Distance between ax + by + c1 = 0 and ax + by + c2 = 0 = |c1 − c2|/√(a² + b²)

Foot of perpendicular and image

(x − x1)/a = (y − y1)/b = −(ax1 + by1 + c)/(a² + b²)   (foot)

For the image, use 2 in place of 1 on the right side: −2(ax1 + by1 + c)/(a² + b²).

Position of points

Two points lie on the same side of ax + by + c = 0 if ax1 + by1 + c and ax2 + by2 + c have the same sign.

Concurrency and family of lines

  • Three lines aix + biy + ci = 0 are concurrent if the 3 × 3 determinant of their coefficients is 0 (and they are not all parallel).
  • Any line through the intersection of L1 = 0 and L2 = 0 is L1 + λL2 = 0.
  • Angle bisectors of a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0: (a1x + b1y + c1)/√(a1² + b1²) = ±(a2x + b2y + c2)/√(a2² + b2²).

5. Formula Master Table

ResultFormula
Slope of ax + by + c = 0−a/b
Angle between linestan θ = |(m1 − m2)/(1 + m1m2)|
Point–line distance|ax1 + by1 + c|/√(a² + b²)
Parallel lines distance|c1 − c2|/√(a² + b²)
Line parallel to ax + by + c = 0ax + by + k = 0
Line perpendicular to ax + by + c = 0bx − ay + k = 0
Family through intersectionL1 + λL2 = 0
Area of triangle formed with axes by x/a + y/b = 1|ab|/2
Image of (x1, y1) in y = x(y1, x1)

6. Mind Map

Straight Lines
Basics
  • Distance, section formula
  • Area, collinearity
  • Triangle centres
Forms
  • Slope, point-slope
  • Intercept, normal
  • Parametric
Relations
  • Angle, parallel, perpendicular
  • Distances
  • Foot, image
Families
  • L1 + λL2 = 0
  • Concurrency
  • Angle bisectors

7. Tips and Tricks

Shortcuts
  • To write a line parallel to ax + by + c = 0, keep a and b and change only c. For a perpendicular line, swap a and b and change one sign.
  • For the image in the line y = x, just swap the coordinates; in y = −x, swap and change both signs.
  • For "line through the intersection of two lines and satisfying a condition", use L1 + λL2 = 0 — no need to find the intersection point.
  • For the orthocentre of a triangle with vertices on the axes, use the fact that the altitudes from the axis vertices are easy to write.
Mistakes students make
  • Using the distance formula for parallel lines without first making the x and y coefficients identical.
  • Forgetting the absolute value in the angle formula and getting an obtuse angle by mistake.
  • Treating a vertical line (x = k) as having slope 0 — its slope is undefined.

8. Important One-liners

  • Three points are collinear if the area of the triangle formed by them is zero.
  • The x-axis is y = 0 and the y-axis is x = 0.
  • Lines y = m1x + c1 and y = m2x + c2 are perpendicular if m1m2 = −1.
  • The perpendicular distance from the origin to ax + by + c = 0 is |c|/√(a² + b²).
  • The medians of a triangle meet at the centroid, which divides each median in the ratio 2 : 1.
  • The locus of a point equidistant from two fixed points is the perpendicular bisector of the segment joining them.
  • The area of a parallelogram formed by lines a1x + b1y = c1, d1 and a2x + b2y = c2, d2 is |(c1 − d1)(c2 − d2)/(a1b2 − a2b1)|.

9. PYQ-pattern Questions with Solutions

These questions are original and modelled on concepts frequently tested in JEE Main.

Q1Numerical

Find the distance of the point (3, 4) from the line 3x + 4y − 5 = 0.

Show solution

Distance = |3(3) + 4(4) − 5|/√(9 + 16) = |9 + 16 − 5|/5 = 20/5 = 4.

Answer: 4

Q2Easy

The distance between the parallel lines 3x + 4y = 7 and 6x + 8y + 6 = 0 is:

  1. 1/5
  2. 2
  3. 13/10
  4. 1
Show solution

Make the coefficients equal: 6x + 8y + 6 = 0 ⇒ 3x + 4y + 3 = 0. The first line is 3x + 4y − 7 = 0.

Distance = |−7 − 3|/√(9 + 16) = 10/5 = 2.

Answer: (B)

Q3Easy

The acute angle between the lines y = √3 x and y = x/√3 is:

  1. 15°
  2. 30°
  3. 45°
  4. 60°
Show solution

The lines make angles 60° and 30° with the x-axis, so the angle between them is 30°.

(Formula: tan θ = |(√3 − 1/√3)/(1 + 1)| = (2/√3)/2 = 1/√3 ⇒ θ = 30°.)

Answer: (B)

Q4Medium

The image of the point (1, 2) in the line x − y = 0 is:

  1. (−1, −2)
  2. (2, 1)
  3. (−2, −1)
  4. (1, −2)
Show solution

The line x − y = 0 is y = x; reflecting in it swaps the coordinates. Formula check: (x − 1)/1 = (y − 2)/(−1) = −2(1 − 2)/2 = 1 ⇒ x = 2, y = 1.

Answer: (B)

Q5Medium

The equation of the line through the intersection of x + y = 3 and x − y = 1, and parallel to x + 2y = 7, is:

  1. x + 2y = 4
  2. x + 2y = 5
  3. 2x + y = 5
  4. x + 2y = 3
Show solution

The intersection of x + y = 3 and x − y = 1 is (2, 1). A line parallel to x + 2y = 7 is x + 2y = k; through (2, 1): k = 2 + 2 = 4.

Answer: (A)

Q6Numerical

Find the area of the triangle formed by the line 3x + 2y = 12 with the coordinate axes.

Show solution

Intercept form: x/4 + y/6 = 1, so the intercepts are 4 and 6.

Area = ½ × 4 × 6 = 12.

Answer: 12

Q7Medium

The foot of the perpendicular from the origin to the line x + y = 2 is:

  1. (2, 0)
  2. (1, 1)
  3. (0, 2)
  4. (½, ½)
Show solution

(x − 0)/1 = (y − 0)/1 = −(0 + 0 − 2)/(1 + 1) = 1 ⇒ foot = (1, 1).

Answer: (B)

Q8Hard

If the lines ax + y + 1 = 0, x + by + 1 = 0 and x + y + c = 0 (a, b, c ≠ 1) are concurrent, then 1/(1 − a) + 1/(1 − b) + 1/(1 − c) equals:

  1. 0
  2. 1
  3. 2
  4. −1
Show solution

Concurrency: |a 1 1; 1 b 1; 1 1 c| = 0. Apply C1 → C1 − C3 and C2 → C2 − C3:

|a − 1 0 1; 0 b − 1 1; 1 − c 1 − c c| = 0. Expanding and dividing by (1 − a)(1 − b)(1 − c) gives

c/(1 − c) + 1/(1 − b) + 1/(1 − a) = 0, and since c/(1 − c) = 1/(1 − c) − 1, we get 1/(1 − a) + 1/(1 − b) + 1/(1 − c) = 1.

Answer: (B)

10. Quick Revision Summary

Remember before the exam
  • Slope −a/b; parallel: same slope; perpendicular: product −1.
  • Distance of a point: |ax1 + by1 + c|/√(a² + b²); parallel lines: |c1 − c2|/√(a² + b²).
  • Foot uses −(…)/(a² + b²); image uses −2(…)/(a² + b²).
  • L1 + λL2 = 0 for lines through an intersection; determinant = 0 for concurrency.
  • Centroid divides HO in 2 : 1 (Euler line).
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