MHT-CET Maths · Pair of Straight Lines

Slopes of a Homogeneous Pair — Sum, Product and Ratio Conditions

For ax² + 2hxy + by² = 0 the slopes satisfy m₁ + m₂ = −2h/b and m₁m₂ = a/b; any relation between the slopes (a ratio, a reciprocal, a common line) becomes an equation in a, h, b.

All Pair of Straight Lines notes MHT-CET Maths strategy

Why this matters

16 PYQs at 56% HARD — the algebraic heart of the chapter and its densest HARD page. 'One slope is k times the other' has been set with k = 2, 3, 4 and the ratio 2 : 3; 'the slopes are reciprocals' once; 'a common line between two pairs' once; and the general identity (m₁ + m₂)²/(m₁m₂) = 4h²/(ab) twice, disguised as 16h² = 25ab and 4ab = 3h². The HARD ones are the same identity with a parameter to eliminate; the trap is forgetting that b, not a, divides the coefficients.

Concept 1 of 3: m₁ + m₂ = −2h/b and m₁m₂ = a/b

Divide

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

x^2

b\left(\dfrac{y}{x}\right)^2 + 2h\left(\dfrac{y}{x}\right) + a = 0

is a quadratic in the slope

m = \dfrac{y}{x}

. Vieta on it gives the sum and the product.

Definition

$Kx^2 + 6xy + y^2 = 0$ : $m_1 + m_2 = -6$ , $m_1 m_2 = K$ . With $m_2 = 3m_1$ : $4m_1 = -6$ , $m_1 = -\tfrac32$ , $K = 3m_1^2 = \tfrac{27}{4}$ .
$4x^2 + kxy + y^2 = 0$ , one slope four times the other: $4m_2^2 = 4$ , $m_2 = \pm1$ , $5m_2 = -k$ , $k = \mp5$ ; the paper offers $5$ .
$x^2 + 2hxy + 2y^2 = 0$ , slopes in $1 : 2$ : $2m^2 = \tfrac12$ , $m = \pm\tfrac12$ ; $3m = -h$ , $h = \mp\tfrac32$ ; $\tfrac32$ is offered.
Reciprocal slopes ( $ax^2 + (2a + 1)xy + 2y^2 = 0$ ): $m_1 m_2 = \tfrac{a}{2} = 1 \Rightarrow a = 2$ ; sum $-\tfrac{5}{2}$ ; $m_1^2 + m_2^2 = \tfrac{25}{4} - 2 = \tfrac{17}{4}$ .
Divide by $b$ , the coefficient of $y^2$ . Dividing by $a$ gives the sum and product of the RECIPROCAL slopes.

Vieta for slopes

bm^2 + 2hm + a = 0:\qquad m_1 + m_2 = -\frac{2h}{b},\qquad m_1 m_2 = \frac{a}{b}

Worked example

One line of

2x^2 + kxy + 3y^2 = 0

has slope

-2

. Find

k

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 11th May Shift 1 · Q142Moderate

Example 1 · Pair of Straight Lines · Slopes of a Homogeneous Pair — Sum, Product and Ratio Conditions

If the slope of one of the lines given by

Kx^2 + 6xy + y^2 = 0

is three times the other, then the value of K is

Dividing by a instead of b

m_1 m_2 = \dfrac{a}{b}

, the

x^2

coefficient over the

y^2

coefficient. Inverting it gives

K = \dfrac{4}{27}

, option (D) on the

Kx^2 + 6xy + y^2

stem.

Drill 3 more on m₁ + m₂ = −2h/b and m₁m₂ = a/b

Concept 2 of 3: Slopes in a Ratio: (m + n)² ab = 4mn h², and the Reverse Direction

m_1 : m_2 = m : n

, write

m_1 = mt

m_2 = nt

and eliminate

t

between the sum and the product:

\dfrac{(m_1 + m_2)^2}{m_1 m_2} = \dfrac{(m + n)^2}{mn} = \dfrac{4h^2}{ab}

. The same identity, read backwards, turns a given relation like

16h^2 = 25ab

into the ratio.

Definition

Ratio $1 : 2$ : $\dfrac{9}{2} = \dfrac{4h^2}{ab} \Rightarrow 9ab = 8h^2$ , so $ab : h^2 = 8 : 9$ .
Ratio $2 : 3$ for $6x^2 + 2hxy + y^2 = 0$ : $25 \cdot 6 = 4 \cdot 6 \cdot h^2 \Rightarrow h^2 = \tfrac{25}{4}$ , $h = \pm\tfrac52$ .
Given $16h^2 = 25ab$ : $\dfrac{(k + 1)^2}{k} = \dfrac{4h^2}{ab} = \dfrac{25}{4} \Rightarrow 4k^2 - 17k + 4 = 0 \Rightarrow k = 4$ or $\tfrac14$ : one slope is four times the other.
Given $4ab = 3h^2$ : $(m_1 - m_2)^2 = \dfrac{4h^2 - 4ab}{b^2} = \dfrac{h^2}{b^2}$ , so $m_1 - m_2 = \pm\dfrac{h}{b}$ with $m_1 + m_2 = -\dfrac{2h}{b}$ : slopes $-\dfrac{h}{2b}$ , $-\dfrac{3h}{2b}$ , ratio $1 : 3$ .
$\dfrac{(m_1 + m_2)^2}{m_1 m_2}$ is the quantity to compute in every ratio stem; it is $\dfrac{4h^2}{ab}$ whatever $b$ is.

Ratio identity

m_1 : m_2 = m : n \iff (m + n)^2\,ab = 4mn\,h^2 \qquad \frac{(m_1 + m_2)^2}{m_1 m_2} = \frac{4h^2}{ab}

Worked example

The slopes of

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

are in the ratio

2 : 5

. Find

ab : h^2

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 19 April Shift I · Q143Hard

Example 2 · Pair of Straight Lines · Slopes of a Homogeneous Pair — Sum, Product and Ratio Conditions

m_{1}

and

m_{2}

are the slopes of the lines represented by

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

satisfying the condition

16{\text{ }h}^{2}= 25ab

, then ....

Writing the identity with h² and ab swapped

(m + n)^2 ab = 4mn h^2

, so

ab : h^2 = 8 : 9

for the ratio

1 : 2

, not

9 : 8

. Both orders are always offered; check with

m = n = 1

, which must give

h^2 = ab

Drill 3 more on slopes in a ratio: (m + n)² ab = 4mn h², and the reverse direction

Concept 3 of 3: A Common Line Between Two Pairs, and a Line of the Pair Perpendicular to a Given Line

A line

y = mx

belongs to a pair iff

m

satisfies its slope quadratic. So a common line means a common root; and 'one line of the pair is perpendicular to

mx + ny = 18

' means the slope

\dfrac{n}{m}

is a root.

Definition

$6x^2 - xy - 5y^2 = (6x + 5y)(x - y)$ : lines $x = y$ and $x = -\tfrac56 y$ . For $3x^2 - 5xy + py^2 = 0$ to share one: $x = y$ gives $3 - 5 + p = 0$ , $p = 2$ ; $x = -\tfrac56 y$ gives $\tfrac{75}{36} + \tfrac{25}{6} + p = 0$ , $p = -\tfrac{25}{4}$ .
One line of $ax^2 + 2hxy + by^2 = 0$ perpendicular to $mx + ny = 18$ (slope $-\tfrac{m}{n}$ ) has slope $\tfrac{n}{m}$ : substitute $y = \tfrac{n}{m}x$ : $am^2 + 2hmn + bn^2 = 0$ .
Substituting a direction $(x, y) = (m, n)$ into the pair is the fastest membership test.
Two pairs sharing BOTH lines are proportional equations; sharing one is a single common root.

Membership test

y = kx \text{ belongs to } ax^2 + 2hxy + by^2 = 0 \iff a + 2hk + bk^2 = 0

Worked example

For what

p

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

and

px^2 - 4xy + y^2 = 0

have a common line?

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 26 April Shift II · Q123Hard

Example 3 · Pair of Straight Lines · Slopes of a Homogeneous Pair — Sum, Product and Ratio Conditions

If the pair of lines

3x^{2}- 5xy+ py^{2}= 0

and

6x^{2}-xy- 5y^{2}= 0

have one line common, then

p =

Substituting the given line's own slope

Perpendicular to

mx + ny = 18

means slope

\dfrac{n}{m}

, which gives

am^2 + 2hmn + bn^2 = 0

. Using

-\dfrac{m}{n}

swaps

m

and

n

and flips a sign — options (A) and (D) on that stem.

Drill 1 more on a common line between two pairs, and a line of the pair perpendicular to a given line

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)

m₁ + m₂ = −2h/b and m₁m₂ = a/b
Vieta for slopes
$bm^2 + 2hm + a = 0:\qquad m_1 + m_2 = -\frac{2h}{b},\qquad m_1 m_2 = \frac{a}{b}$
Slopes in a Ratio: (m + n)² ab = 4mn h², and the Reverse Direction
Ratio identity
$m_1 : m_2 = m : n \iff (m + n)^2\,ab = 4mn\,h^2 \qquad \frac{(m_1 + m_2)^2}{m_1 m_2} = \frac{4h^2}{ab}$
A Common Line Between Two Pairs, and a Line of the Pair Perpendicular to a Given Line
Membership test
$y = kx \text{ belongs to } ax^2 + 2hxy + by^2 = 0 \iff a + 2hk + bk^2 = 0$

Watch out for (3)

Dividing by a instead of b
→ m₁ + m₂ = −2h/b and m₁m₂ = a/b
Writing the identity with h² and ab swapped
→ Slopes in a Ratio: (m + n)² ab = 4mn h², and the Reverse Direction
Substituting the given line's own slope
→ A Common Line Between Two Pairs, and a Line of the Pair Perpendicular to a Given Line

Test yourself on Pair of Straight Lines

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