MHT-CET Maths · Pair of Straight Lines

Angle Between the Pair — Perpendicular Pairs, Lines at a Given Angle and the Bisectors

tan θ = 2√(h² − ab)/|a + b| is the angle between the two lines of ax² + 2hxy + by² = 0; a + b = 0 means perpendicular, h² = ab parallel, and the bisectors are (x² − y²)/(a − b) = xy/h.

All Pair of Straight Lines notes MHT-CET Maths strategy

Why this matters

21 PYQs at 48% HARD — the page with the most HARD questions in the chapter. The stems are lines through a point at 45° or 60° to a given line written as a joint equation (four sittings), the perpendicular condition a + b = 0 in a trigonometric disguise, the pair perpendicular to a given pair through a point, the angle equal to 2θ with a parameter, a right isosceles triangle's two legs (twice), and the bisector pair (twice). The pair through a point at a given angle is the Straight Line chapter's two-root problem multiplied out — that is where the marks are.

Concept 1 of 3: tan θ = 2√(h² − ab)/|a + b|: Perpendicular When a + b = 0, Parallel When h² = ab

Feed

m_1 + m_2

and

m_1 m_2

into

\tan\theta = \left|\dfrac{m_1 - m_2}{1 + m_1 m_2}\right|

: the numerator becomes

\dfrac{2\sqrt{h^2 - ab}}{|b|}

and the denominator

\dfrac{a + b}{b}

. The angle is

90^\circ

exactly when

a + b = 0

, and

0

when

h^2 = ab

Definition

$(x\cos\alpha + y\sin\alpha)^2 = (x^2 + y^2)\sin^2\alpha$ : expand and collect, $x^2(\cos^2\alpha - \sin^2\alpha) + xy\sin 2\alpha = 0$ , i.e. $x^2\cos 2\alpha + xy\sin 2\alpha = 0$ ; $a = \cos 2\alpha$ , $b = 0$ ; perpendicular iff $a + b = 0$ , so $\cos 2\alpha = 0$ , $\alpha = \dfrac{\pi}{4}$ — set in two 2024 shifts.\n- $x^2 + \lambda xy - y^2\tan^2\theta = 0$ with angle $2\theta$ : $\tan 2\theta = \dfrac{2\sqrt{\lambda^2/4 + \tan^2\theta}}{1 - \tan^2\theta}$ and $\tan 2\theta = \dfrac{2\tan\theta}{1 - \tan^2\theta}$ force $\lambda = 0$ .
Diagonals along $x + 3y = 4$ and $6x - 2y = 7$ have slopes $-\tfrac13$ and $3$ : perpendicular, so the parallelogram is a rhombus.
The formula returns the acute angle; $h^2 - ab < 0$ means no real lines at all.

Angle between the pair

\tan\theta = \frac{2\sqrt{h^2 - ab}}{|a + b|} \qquad a + b = 0 \iff \perp \qquad h^2 = ab \iff \parallel

Worked example

Find the acute angle between the lines

x^2 + 4xy + y^2 = 0

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 12th May Shift 2 · Q117Moderate

Example 1 · Pair of Straight Lines · Angle Between the Pair — Perpendicular Pairs, Lines at a Given Angle and the Bisectors

If the pair of lines given by

(x\cos\alpha + y\sin\alpha)^2 = (x^2+y^2)\sin^2\alpha

are perpendicular to each other, then

\alpha

Using |a − b| in the denominator

The denominator is

|a + b|

;

a - b

belongs to the BISECTOR formula. Mixing them turns a

60^\circ

answer into

\tan^{-1}

of something not on the list — or worse, onto a distractor.

Drill 2 more on tan θ = 2√(h² − ab)/|a + b|: perpendicular when a + b = 0, parallel when h² = ab

Concept 2 of 3: The Pair Through a Point at a Given Angle to a Line: Square the Angle Condition

Both lines through the origin at angle

\alpha

y = m_1 x

satisfy

\left(\dfrac{m - m_1}{1 + m m_1}\right)^2 = \tan^2\alpha

. Put

m = \dfrac{y}{x}

, clear denominators, and the squared condition IS the joint equation — both roots at once, no need to find them.

Definition

At $45^\circ$ to $3x + y = 0$ ( $m_1 = -3$ ): $(m + 3)^2 = (1 - 3m)^2 \Rightarrow 2m^2 - 3m - 2 = 0$ ; with $m = \dfrac{y}{x}$ : $2y^2 - 3xy - 2x^2 = 0$ , i.e. $2x^2 + 3xy - 2y^2 = 0$ .
At $\dfrac{\pi}{4}$ to $3x + 2y - 8 = 0$ ( $m_1 = -\tfrac32$ ): $5m^2 - 24m - 5 = 0 \Rightarrow 5x^2 + 24xy - 5y^2 = 0$ .
At $\dfrac{\pi}{6}$ to $3x + y - 6 = 0$ ( $m_1 = -3$ ): $3(m + 3)^2 = (1 - 3m)^2 \Rightarrow 3m^2 - 12m - 13 = 0 \Rightarrow 13x^2 + 12xy - 3y^2 = 0$ .
Closed form: lines through the origin at angle $\alpha$ to $ax + by = 0$ are $(ax + by)^2 = \tan^2\alpha\,(bx - ay)^2$ .
Pair through $(3, -2)$ perpendicular to $5x^2 + 2xy - 3y^2 = 0$ : the given slopes are $\tfrac53$ and $-1$ , the perpendicular slopes $-\tfrac35$ and $1$ ; lines $3x + 5y + 1 = 0$ and $x - y - 5 = 0$ , product $3x^2 + 2xy - 5y^2 - 14x - 26y - 5 = 0$ . Shortcut for the homogeneous part: swap $a$ and $b$ and change the sign of $h$ .
Right isosceles triangle at $Q(2, 1)$ with hypotenuse on $2x + y = 3$ : the legs make $45^\circ$ with slope $-2$ , slopes $-\tfrac13$ and $3$ ; through $Q$ : $(x + 3y - 5)(3x - y - 5) = 3x^2 + 8xy - 3y^2 - 20x - 10y + 25 = 0$ .

Pair at angle α to ax + by = 0

(ax + by)^2 = \tan^2\alpha\,(bx - ay)^2 \qquad \text{perpendicular pair to } ax^2 + 2hxy + by^2: \ bx^2 - 2hxy + ay^2 = 0

Worked example

Find the joint equation of the lines through the origin making

60^\circ

with

x + y = 0

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 10th May Shift 2 · Q104Hard

Example 2 · Pair of Straight Lines · Angle Between the Pair — Perpendicular Pairs, Lines at a Given Angle and the Bisectors

The joint equation of a pair of lines passing through the origin and making an angle of

\frac{\pi}{4}

with the line

3x+2y-8=0

Sign of the xy term after substituting m = y/x

5m^2 - 24m - 5 = 0

becomes

5y^2 - 24xy - 5x^2 = 0

; multiplying by

-1

gives

5x^2 + 24xy - 5y^2 = 0

. The three wrong options differ only in these signs; keep the substitution explicit.

Drill 5 more on the pair through a point at a given angle to a line: square the angle condition

Concept 3 of 3: The Angle Bisectors: (x² − y²)/(a − b) = xy/h

The two bisectors of the angles between the lines of

ax^2 + 2hxy + by^2 = 0

form their own pair,

h(x^2 - y^2) = (a - b)xy

. It is always a perpendicular pair (the coefficients of

x^2

and

y^2

are

h

and

-h

Definition

$x^2 - 4xy - 5y^2 = 0$ : $a = 1$ , $b = -5$ , $h = -2$ : $\dfrac{x^2 - y^2}{6} = \dfrac{xy}{-2} \Rightarrow x^2 + 3xy - y^2 = 0$ .
$2x^2 + 11xy + 3y^2 = 0$ : $a - b = -1$ , $h = \tfrac{11}{2}$ : $-(x^2 - y^2) = \dfrac{2xy}{11} \Rightarrow 11x^2 + 2xy - 11y^2 = 0$ .
$2h$ is the coefficient of $xy$ , so $h$ is HALF of it; forgetting the half is the standard error here.
Bisectors of $x = 5$ and $y = 3$ (not through the origin) come from slopes $\pm1$ at the intersection, page 1 — the formula above is for pairs through the origin.

Bisector pair

\frac{x^2 - y^2}{a - b} = \frac{xy}{h}

Worked example

Find the bisectors of the angles between the lines

3x^2 - 8xy - 3y^2 = 0

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 3rd May Shift 2 · Q106Hard

Example 3 · Pair of Straight Lines · Angle Between the Pair — Perpendicular Pairs, Lines at a Given Angle and the Bisectors

The equation of pair of lines

y=px

and

y=qx

can be written as

(y-px)(y-qx)=0

. Then the equation of the pair of the angle bisectors of the lines

x^{2}-4xy-5y^{2}=0

Using the full xy coefficient as h

For

2x^2 + 11xy + 3y^2 = 0

h = \dfrac{11}{2}

. Using

11

gives

11x^2 + 4xy - 11y^2

-type answers that miss every option by a factor in the middle term.

Drill 1 more on the angle bisectors: (x² − y²)/(a − b) = xy/h

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)

tan θ = 2√(h² − ab)/|a + b|: Perpendicular When a + b = 0, Parallel When h² = ab
Angle between the pair
$\tan\theta = \frac{2\sqrt{h^2 - ab}}{|a + b|} \qquad a + b = 0 \iff \perp \qquad h^2 = ab \iff \parallel$
The Pair Through a Point at a Given Angle to a Line: Square the Angle Condition
Pair at angle α to ax + by = 0
$(ax + by)^2 = \tan^2\alpha\,(bx - ay)^2 \qquad \text{perpendicular pair to } ax^2 + 2hxy + by^2: \ bx^2 - 2hxy + ay^2 = 0$
The Angle Bisectors: (x² − y²)/(a − b) = xy/h
Bisector pair
$\frac{x^2 - y^2}{a - b} = \frac{xy}{h}$

Watch out for (3)

Using |a − b| in the denominator
→ tan θ = 2√(h² − ab)/|a + b|: Perpendicular When a + b = 0, Parallel When h² = ab
Sign of the xy term after substituting m = y/x
→ The Pair Through a Point at a Given Angle to a Line: Square the Angle Condition
Using the full xy coefficient as h
→ The Angle Bisectors: (x² − y²)/(a − b) = xy/h

Test yourself on Pair of Straight Lines

15 past MHT-CET questions from this chapter, timed at 27 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Pair of Straight Lines chapter test Drill all 21 questions

Related notes

Straight Line — the angle formula this page multiplies out →