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

12 PYQs at 58% HARD — the chapter's densest HARD page. 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 (twice), 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

Intuition

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

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1Pair of Straight LinesMODERATE

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

[Q117 · 13th May Shift 1 · 2024]

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 3 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

Intuition

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

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Pair of Straight LinesHARD

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

[Q104 · 10th May Shift 2 · 2024]

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

Intuition

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

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3Pair of Straight LinesHARD

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

[Q106 · 3rd May Shift 2 · 2023]

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

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