Straight Lines
NCERT Class 11 Mathematics Chapter 9
Straight Lines for Class 11 Maths and JEE: slope, angle between lines, the forms of the equation of a line and the distance of a point from a line.
6 topics Updated 8 Nov 2026
For: Class 11 and JEE, unless a topic says otherwise
In the syllabus: JEE Class 11 syllabus
With your textbook
NCERT Mathematics (Class 11), Chapter 9 “Straight Lines”, reprint 2026-27.
Open the official NCERT chapter (PDF, 2.5 MB, ncert.nic.in) (opens in a new tab)
Checked on 19 Oct 2026.
Learn in this order
-
Coordinate geometry basics
Proofs, external division, mid-point, triangle area, collinear points and solved examples.
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Locus and its equation
Locus of a point for JEE Main: the P(h, k) method, perpendicular bisectors, circles, angle bisectors and loci with a parameter, with solved examples.
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Slope of a line
Inclination, m = tan θ, slope from two points, parallel and perpendicular lines and the angle between two lines.
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Forms of the equation of a line
Point-slope, two-point, slope-intercept, intercept and general forms, a formula sheet and solved NCERT questions.
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Distance of a point from a line
The formula and its proof, parallel lines, the foot of the perpendicular and the image of a point.
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Intersection of lines and centres of a triangle
Intersection of two lines for JEE Main: the formula, concurrent lines, and the centroid, circumcentre and orthocentre of a triangle, with solved examples.
Key formulas in this chapter
Coordinate geometry basics
- PQ = (x − x) + (y − y)
- internal m: n: ((mx + nx)/(m + n), (my + ny)/(m + n))
- external m: n: ((mx − nx)/(m − n), (my − ny)/(m − n)), m, n > 0, m ≠ n
- mid-point: ((x + x)/2, (y + y)/2)
- area = ½|x(y − y) + x(y − y) + x(y − y)|
- collinear ⇔ x(y − y) + x(y − y) + x(y − y) = 0
- polygon: ½|(xy − xy) + … + (xy − xy)|
Locus
- P(h, k) on the locus → condition in h, k → replace h, k by x, y
- PA = PB ⇔ P on the perpendicular bisector of AB (A ≠ B)
- PC = r: (x − a) + (y − b) = r, r ≥ 0
- rod of length l sliding on the axes, mid-point: x + y = l/4
Slope of a line
- m = tan θ, θ ≠ 90°
- m = (y − y)/(x − x), x ≠ x
- distinct non-vertical lines parallel: m = m
- perpendicular: mm = −1
- tan θ = |(m − m)/(1 + mm)|, 1 + mm ≠ 0
- A, B, C collinear ⇔ slope of AB = slope of BC (AB vertical: equal x-coordinates)
Forms of the equation of a line
- horizontal: y = a; vertical: x = b
- y − y = m(x − x)
- y − y = ((y − y)/(x − x))(x − x), x ≠ x
- y = mx + c; y = m(x − d)
- x/a + y/b = 1 (a, b ≠ 0)
- Ax + By + C = 0: slope −A/B (B ≠ 0); intercepts −C/A (A ≠ 0), −C/B (B ≠ 0)
- parallel through (x, y): A(x − x) + B(y − y) = 0
- perpendicular through (x, y): B(x − x) − A(y − y) = 0
Distance of a point from a line
- d = |Ax + By + C| / √(A + B)
- origin: d = |C| / √(A + B)
- parallel lines: d = |C − C| / √(A + B)
- y = mx + c, y = mx + c: d = |c − c| / √(1 + m)
- foot: (h − x)/A = (k − y)/B = −(Ax + By + C)/(A + B), A, B ≠ 0
- image: (h − x)/A = (k − y)/B = −2(Ax + By + C)/(A + B), A, B ≠ 0
Intersection of lines and triangle centres
- D = ab − ab ≠ 0: x = (bc − bc)/D, y = (ca − ca)/D
- parallel or the same line: ab − ab = 0
- concurrent (no two parallel): a(bc − bc) + b(ca − ca) + c(ab − ab) = 0
- through the intersection of L = 0 and L = 0: L + λL = 0
- G = ((x + x + x)/3, (y + y + y)/3)
- circumcentre O: OA = OB = OC
- orthocentre H: meet of the altitudes; H = 3G − 2O
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