MHT-CET Maths · Pair of Straight Lines

Joint Equation of Two Lines — Product of Linear Factors and the Triangle They Form

Multiply two linear equations to get the joint equation; factorise a joint equation to get the two lines back — and with a third line, the three lines bound a triangle whose vertices, centroid and circumcentre follow from the intersections.

All Pair of Straight Lines notes MHT-CET Maths strategy

Why this matters

17 PYQs at 18% HARD — the cheapest page in the chapter. Lines through the origin at 30° to the Y-axis (set twice), lines through a point parallel to the axis bisectors, the median-and-altitude pair from a vertex, two normal-form lines multiplied, and — the 2025 favourite — a factorable pair plus a third line forming a right triangle whose circumcentre or circumradius is asked. Two of the three HARD ones factorise a pair and then do triangle geometry; nothing here is beyond expanding a product.

Concept 1 of 3: The Joint Equation Is the Product: (L₁)(L₂) = 0

A point lies on line

L_1

or line

L_2

exactly when

L_1 \cdot L_2 = 0

. So the pair is written by multiplying, and for lines through the origin the result is homogeneous:

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

Definition

Lines through the origin at $30^\circ$ to the $Y$ -axis make $60^\circ$ with the $X$ -axis: slopes $\pm\sqrt3$ ; $(y - \sqrt3 x)(y + \sqrt3 x) = 0 \Rightarrow 3x^2 - y^2 = 0$ . Lines forming an equilateral triangle with $y = 5$ are the same pair.
Through $(-2, 3)$ parallel to the axis bisectors (slopes $\pm1$ ): $(x - y + 5)(x + y - 1) = x^2 - y^2 + 4x + 6y - 5 = 0$ .
Bisectors of the angles between $x = 5$ and $y = 3$ : through $(5, 3)$ with slopes $\pm1$ : $(x - y - 2)(x + y - 8) = x^2 - y^2 - 10x + 6y + 16 = 0$ .
Normal-form lines at unit distance with normals at $\tfrac{\pi}{4}$ and $\tfrac{3\pi}{4}$ : $x + y = \sqrt2$ and $-x + y = \sqrt2$ ; product $x^2 - y^2 + 2\sqrt2 y - 2 = 0$ .
Median and altitude from $O$ in $O(0,0)$ , $A(1,2)$ , $B(3,4)$ : median to $(2, 3)$ is $3x - 2y = 0$ , altitude perpendicular to $AB$ (slope $1$ ) is $x + y = 0$ : $3x^2 + xy - 2y^2 = 0$ .

Joint equation

(a_1x + b_1y + c_1)(a_2x + b_2y + c_2) = 0 \qquad \text{through the origin: } (y - m_1x)(y - m_2x) = 0

Worked example

Write the joint equation of the lines through

(1, 2)

with slopes

2

and

-3

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 11th May Shift 1 · Q139Moderate

Example 1 · Pair of Straight Lines · Joint Equation of Two Lines — Product of Linear Factors and the Triangle They Form

The joint equation of two lines through the origin, each making an angle with measure of

30^\circ

with the positive Y-axis, is

30° to the Y-axis read as slope tan 30°

A line at

30^\circ

to the

Y

-axis is at

60^\circ

to the

X

-axis: slope

\sqrt3

, joint equation

3x^2 - y^2 = 0

. Option

x^2 - 3y^2 = 0

is the

\tan 30^\circ

slip.

Drill 6 more on the joint equation is the product: (l₁)(l₂) = 0

Concept 2 of 3: Factorising a Pair: Split the Middle Term, or Complete the Square

A homogeneous pair factorises like a quadratic in

\dfrac{y}{x}

; a non-homogeneous pair often splits by grouping (

xy - x + y - 1 = (x + 1)(y - 1)

) or as a difference of squares after completing squares.

Definition

$6x^2 + xy - y^2 = (2x + y)(3x - y)$ ; $2x^2 - 5xy + 2y^2 = (2x - y)(x - 2y)$ ; $x^2 - 4xy - 5y^2 = (x - 5y)(x + y)$ .
$xy - x + y - 1 = (x + 1)(y - 1)$ : lines $x = -1$ , $y = 1$ , meeting at $(-1, 1)$ . A third line $x + ky - 3 = 0$ is concurrent with them iff it passes through that point: $k = 4$ .
$xy + 2x + 2y + 4 = (x + 2)(y + 2)$ : the lines $x = -2$ , $y = -2$ .
$x^2 - y^2 - 2x + 4y - 3 = (x - 1)^2 - (y - 2)^2 = (x + y - 3)(x - y + 1)$ — complete both squares, then difference of squares.

Factorising

ax^2 + 2hxy + by^2 = b(y - m_1x)(y - m_2x) \qquad \text{grouping: } xy + px + qy + pq = (x + q)(y + p)

Worked example

Factorise

3x^2 + 7xy + 2y^2 = 0

into two lines.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 22 April Shift I · Q119Easy

Example 2 · Pair of Straight Lines · Joint Equation of Two Lines — Product of Linear Factors and the Triangle They Form

If the pair of straight lines

xy-x+y- 1 = 0

and the line

x+ky- 3 = 0

are concurrent, then the value of

k

is equal to

Reading xy − x + y − 1 as a curve

Any equation of the form

xy + px + qy + pq = 0

is two lines. Grouping finds them; treating it as a hyperbola sends you down the wrong chapter.

Concept 3 of 3: The Triangle a Pair Makes With a Third Line: Vertices, Centroid, Median, Circumcentre

Factorise the pair, intersect each factor with the third line, and the three vertices are known — the pair's own intersection is the third. Then the centroid is the average, the median joins a vertex to a midpoint, and a right angle puts the circumcentre at the midpoint of the hypotenuse.

Definition

$6x^2 + xy - y^2 = 0$ with $x + 3y = 10$ : lines $2x + y = 0$ , $3x - y = 0$ ; vertices $(0, 0)$ , $(-2, 4)$ , $(1, 3)$ ; centroid $\left(-\tfrac13, \tfrac73\right)$ .
$x^2 - 4xy + y^2 = 0$ with $AB: 2x + 3y = 1$ : substitute $x = \dfrac{1 - 3y}{2}$ to get $37y^2 - 14y + 1 = 0$ ; the midpoint of $AB$ has $y = \dfrac{7}{37}$ (half the sum of roots) and $x = \dfrac{8}{37}$ ; the median from $O$ is $7x - 8y = 0$ . No need to find $A$ and $B$ themselves.
$xy + 2x + 2y + 4 = 0$ ( $x = -2$ , $y = -2$ ) with $x + y + 2 = 0$ : vertices $(-2, -2)$ , $(0, -2)$ , $(-2, 0)$ , right angle at $(-2, -2)$ ; circumcentre is the midpoint of the hypotenuse $(-1, -1)$ , circumradius $\dfrac{2\sqrt2}{2} = \sqrt2$ .
The midpoint-by-Vieta trick works for any pair through the origin: the two intersections with a line are the roots of one quadratic.

Triangle from a pair

\text{vertices: } O,\ L_1 \cap L_3,\ L_2 \cap L_3;\qquad \text{centroid} = \frac{\sum \text{vertices}}{3}

Worked example

Find the centroid of the triangle formed by

x^2 - y^2 = 0

and

x = 3

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 14th May Shift 2 · Q129Hard

Example 3 · Pair of Straight Lines · Joint Equation of Two Lines — Product of Linear Factors and the Triangle They Form

The centroid of the triangle formed by the lines

x+3y=10

and

6x^2+xy-y^2=0

Finding A and B explicitly for the median

The median from

O

needs only the MIDPOINT of

AB

, which Vieta gives from the quadratic without solving it. Solving

37y^2 - 14y + 1 = 0

by formula wastes the question's time budget.

Drill 3 more on the triangle a pair makes with a third line: vertices, centroid, median, circumcentre

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 Joint Equation Is the Product: (L₁)(L₂) = 0
Joint equation
$(a_1x + b_1y + c_1)(a_2x + b_2y + c_2) = 0 \qquad \text{through the origin: } (y - m_1x)(y - m_2x) = 0$
Factorising a Pair: Split the Middle Term, or Complete the Square
Factorising
$ax^2 + 2hxy + by^2 = b(y - m_1x)(y - m_2x) \qquad \text{grouping: } xy + px + qy + pq = (x + q)(y + p)$
The Triangle a Pair Makes With a Third Line: Vertices, Centroid, Median, Circumcentre
Triangle from a pair
$\text{vertices: } O,\ L_1 \cap L_3,\ L_2 \cap L_3;\qquad \text{centroid} = \frac{\sum \text{vertices}}{3}$

Watch out for (3)

30° to the Y-axis read as slope tan 30°
→ The Joint Equation Is the Product: (L₁)(L₂) = 0
Reading xy − x + y − 1 as a curve
→ Factorising a Pair: Split the Middle Term, or Complete the Square
Finding A and B explicitly for the median
→ The Triangle a Pair Makes With a Third Line: Vertices, Centroid, Median, Circumcentre

Test yourself on Pair of Straight Lines

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Related notes

Straight Line — section formula and rectangle centres used in the triangle stems →