PYQ Vault

JEE Mains Maths · Straight Lines

Slope, Angle and Forms of a Line

The slope of a line, the angle between two lines, the intercept and normal forms of a line, and the parametric form for measuring distance along a line.

Why this matters

Twenty-one PYQs, all of them multiple choice, and two from 2026. Nine turn on slopes and the angle between two lines: isosceles triangles, lines at a given angle, segments that subtend a given angle at the origin. Eight write a line in intercept or normal form or find where two lines meet, three of them to minimise an area or a sum of intercepts. Four measure a distance along a line in a given direction. Three ideas cover the page.

Concept 1 of 3: The angle between two lines

The slope m=tan⁡θm=\tan\theta is the tangent of the angle a line makes with the positive x-axis. Two lines with slopes m1,m2m_1,m_2 meet at an angle whose tangent is ∣m1−m21+m1m2∣\left|\frac{m_1-m_2}{1+m_1m_2}\right|. Run it backwards to find a line at a given angle to another: the modulus gives two answers, one on each side. Parallel lines have equal slopes; perpendicular lines have m1m2=−1m_1m_2=-1.

Definition

  • Slope through (x1,y1),(x2,y2)(x_1,y_1),(x_2,y_2): m=y2−y1x2−x1m=\frac{y_2-y_1}{x_2-x_1}; slope of ax+by+c=0ax+by+c=0: −ab-\frac ab.
  • Acute angle: tan⁡θ=∣m1−m21+m1m2∣\tan\theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right|.
  • Parallel: m1=m2m_1=m_2; perpendicular: m1m2=−1m_1m_2=-1.
  • Isosceles triangle with its equal sides on two given lines: the base makes equal angles with both, so it is perpendicular to one of their bisectors.

Angle between two lines

tan⁡θ=∣m1−m21+m1m2∣\tan\theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right|

Worked example

Find the slopes of the lines that make 45∘45^\circ with y=3xy=3x.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 28 Jan 2025 · Q142Moderate

Example 1 · Straight Lines · Slope, Angle and Forms of a Line

Two equal sides of an isosceles triangle are along −x+2y=4-x+ 2y= 4 and x+y=4x+y= 4. If mm is the slope of its third side, then the sum of all possible distinct values of mm, is :

Two lines, not one

tan⁡θ=k\tan\theta=k with a modulus gives two slopes. An isosceles triangle with two sides given has two possible bases, and a line at a given angle to another has two positions. Keep both before you add or choose.

Concept 2 of 3: Intercept and normal forms

A line cutting the axes at (a,0)(a,0) and (0,b)(0,b) is xa+yb=1\frac xa+\frac yb=1. The normal form xcos⁡α+ysin⁡α=px\cos\alpha+y\sin\alpha=p uses the length pp of the perpendicular from the origin and the angle α\alpha it makes with the x-axis. For a line through a fixed point (h,k)(h,k) that meets the positive axes, the triangle it cuts off has least area 2hk2hk, and the sum of the intercepts is least at (h+k)2(\sqrt h+\sqrt k)^2.

Definition

  • Intercept form: xa+yb=1\frac xa+\frac yb=1.
  • Normal form: xcos⁡α+ysin⁡α=px\cos\alpha+y\sin\alpha=p, with p≥0p\ge0 the distance from the origin.
  • Through (h,k)(h,k), positive intercepts: least area 2hk2hk; least a+b=(h+k)2a+b=(\sqrt h+\sqrt k)^2.
  • Two lines meet where both equations hold.
  • A line through the centre of a rectangle halves its area.

Intercept and normal forms

xa+yb=1,xcos⁡α+ysin⁡α=p\frac xa+\frac yb=1,\qquad x\cos\alpha+y\sin\alpha=p

Worked example

The perpendicular from the origin to a line has length 4 and makes 60∘60^\circ with the positive x-axis. Find the intercepts.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 23 Jan 2026 Shift 1 · Q54Moderate

Example 2 · Straight Lines · Slope, Angle and Forms of a Line

A rectangle is formed by the lines x=0,y=0x = 0,y = 0, x=3x = 3 and y=4y = 4. Let the line L be perpendicular to 3x+y+6=03x + y + 6 = 0 and divide the area of the rectangle into two equal parts. Then the distance of the point (12,−5)\left( \frac{1}{2}, - 5 \right) from the line L is equal to :

Normal form needs a unit normal

3x+4y=103x+4y=10 is not in normal form until you divide by 32+42=5\sqrt{3^2+4^2}=5: 35x+45y=2\frac35x+\frac45y=2, so p=2p=2. Reading p=10p=10 from the raw equation is the usual slip.

Concept 3 of 3: Distance along a line

Every point on the line through (x1,y1)(x_1,y_1) at angle θ\theta to the x-axis is (x1+rcos⁡θ, y1+rsin⁡θ)(x_1+r\cos\theta,\ y_1+r\sin\theta), and ∣r∣|r| is its distance from (x1,y1)(x_1,y_1). To measure a distance in a given direction, substitute this point into the second line and solve for rr. To find the point at a given distance, put in rr.

Definition

  • x=x1+rcos⁡θx=x_1+r\cos\theta, y=y1+rsin⁡θy=y_1+r\sin\theta.
  • ∣r∣|r| is the distance from (x1,y1)(x_1,y_1); the sign of rr gives the side.
  • "Measured parallel to a line" means along that line's direction.

Parametric form

x−x1cos⁡θ=y−y1sin⁡θ=r\frac{x-x_1}{\cos\theta}=\frac{y-y_1}{\sin\theta}=r

Worked example

Find the distance of (1,2)(1,2) from 2x+y=102x+y=10, measured along a line at 45∘45^\circ to the x-axis.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 24 Jan 2026 Shift 2 · Q53Moderate

Example 3 · Straight Lines · Slope, Angle and Forms of a Line

Let the angles made with the positive x -axis by two straight lines drawn from the point P(2,3)P(2,3) and meeting the line x+y=6x + y = 6 at a distance 23\sqrt{\frac{2}{3}} from the point P be θ1\theta_{1} and θ2\theta_{2}. Then the value of (θ1+θ2)\left( \theta_{1}+\theta_{2} \right) is :

Two directions give two answers

When only the distance is given, θ\theta is the unknown, and the equation for it usually has two solutions: two lines from the point reach the line at that distance. The question may want both, or their sum.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

  • The angle between two lines

    Angle between two lines

    tan⁡θ=∣m1−m21+m1m2∣\tan\theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right|
  • Intercept and normal forms

    Intercept and normal forms

    xa+yb=1,xcos⁡α+ysin⁡α=p\frac xa+\frac yb=1,\qquad x\cos\alpha+y\sin\alpha=p
  • Distance along a line

    Parametric form

    x−x1cos⁡θ=y−y1sin⁡θ=r\frac{x-x_1}{\cos\theta}=\frac{y-y_1}{\sin\theta}=r

Watch out for (3)

Test yourself on Straight Lines

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.