MHT-CET Maths · Pair of Straight Lines

General Second-Degree Equation — Condition for a Pair, Parallel Lines and Distances

ax² + 2hxy + by² + 2gx + 2fy + c = 0 is a pair of lines iff abc + 2fgh − af² − bg² − ch² = 0 (with h² ≥ ab); when h² = ab the pair is parallel and the gap is 2√((g² − ac)/(a(a + b))).

All Pair of Straight Lines notes MHT-CET Maths strategy

Why this matters

10 PYQs at 40% HARD. The determinant condition appears with a parameter to find (k in kxy + 10x + 8y + 16 = 0; the fg = ch identity; a count of integer p), and the parallel-lines case appears as a distance to compute or a p² + q² − pq to evaluate; two stems are the product of perpendicular distances from a point to a homogeneous pair. Read 2g, 2f, 2h off the equation as HALVES — every wrong answer here is a factor of 2 in g, f or h.

Concept 1 of 3

Condition for a Pair: abc + 2fgh − af² − bg² − ch² = 0

Intuition

A second-degree equation factorises into two linear ones only when its

3 \times 3

coefficient determinant vanishes. Identify

a, h, b, g, f, c

with the halves, substitute, and solve for the unknown.

Definition

$kxy + 10x + 8y + 16 = 0$ : $a = b = 0$ , $h = \tfrac{k}{2}$ , $g = 5$ , $f = 4$ , $c = 16$ : $2 \cdot 4 \cdot 5 \cdot \tfrac{k}{2} - 16 \cdot \tfrac{k^2}{4} = 20k - 4k^2 = 0 \Rightarrow k = 0$ or $5$ . $k = 0$ leaves a single line, so $k = 5$ only.
$hxy + gx + fy + c = 0$ : the condition reduces to $\dfrac{fgh}{4} - \dfrac{ch^2}{4} = 0$ , i.e. $fg = ch$ .
$2x^2 + 4xy - py^2 + 4x + qy + 1 = 0$ : the condition gives $p = \dfrac{(q - 4)^2}{4} - 2 \ge -2$ , every $p \ge -2$ attained; integers in $[-5, 5]$ : $-2, \dots, 5$ , eight.
Also required: $h^2 \ge ab$ for the lines to be real. Check it when the answer is a count.

Pair condition

\Delta = \begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix} = abc + 2fgh - af^2 - bg^2 - ch^2 = 0

Worked example

Find

\lambda

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

represents a pair of lines.

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1Pair of Straight LinesMODERATE

If the equation

kxy+ 10x+ 8y+ 16 = 0

represents a pair of lines, then

[Q114 · 20 April Shift II · 2025]

Counting k = 0 as a pair

k = 0

satisfies

\Delta = 0

but reduces the equation to

10x + 8y + 16 = 0

, a single line. 'k = 0 or 5' is option (C) and wrong.

Drill 2 more on condition for a pair: abc + 2fgh − af² − bg² − ch² = 0

Concept 2 of 3

Parallel Pair (h² = ab): Factor as a Perfect Square, or Use 2√((g² − ac)/(a(a + b)))

Intuition

When

h^2 = ab

the quadratic part is a perfect square

(\sqrt a x + \sqrt b y)^2

, and the whole equation is

(\sqrt a x + \sqrt b y + \alpha)(\sqrt a x + \sqrt b y + \beta)

: two parallel lines whose gap is the parallel-lines distance.

Definition

$4x^2 + 4xy + y^2 - 6x - 3y - 4 = (2x + y)^2 - 3(2x + y) - 4 = (2x + y - 4)(2x + y + 1)$ : gap $\dfrac{5}{\sqrt5} = \sqrt5$ .
$(x - 2y + 1)^2 + k(x - 2y + 1) = 0$ : lines $x - 2y + 1 = 0$ and $x - 2y + 1 + k = 0$ , gap $\dfrac{|k|}{\sqrt5} = \sqrt5 \Rightarrow k = 5$ .
$16x^2 - 24xy + 9y^2 + 48x - 36y + 35 = 0$ : $a = 16$ , $g = 24$ , $c = 35$ , $b = 9$ : $2\sqrt{\dfrac{576 - 560}{16 \cdot 25}} = 2 \cdot \dfrac{4}{20} = \dfrac25$ .
$x^2 + 4xy + py^2 + 3x + qy - 4 = 0$ parallel: $p = 4$ ; $(x + 2y)^2 + 3(x + 2y) - 4 = (x + 2y + 4)(x + 2y - 1)$ needs $q = 6$ ; gap $\dfrac{5}{\sqrt5} = \sqrt5$ , $\lambda^2 = 5$ .
$7x^2 - 14xy + py^2 - 12x + qy - 4 = 0$ parallel: $p = 7$ ; $7(x - y)^2 - 12(x - y) - 4 = 0$ forces the $y$ coefficient $q = 12$ ; $p^2 + q^2 - pq = 49 + 144 - 84 = 109$ .

Parallel pair

h^2 = ab:\quad d = 2\sqrt{\frac{g^2 - ac}{a(a + b)}} = 2\sqrt{\frac{f^2 - bc}{b(a + b)}}

Worked example

Find the distance between the lines

x^2 + 2xy + y^2 - 5x - 5y + 6 = 0

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Pair of Straight LinesMODERATE

The distance between the lines represented by the equation

4x^{2}+ 4xy+y^{2}- 6x- 3y- 4 = 0

[Q137 · 20 April Shift I · 2025]

Reporting the gap squared, or halving it

The formula already carries the factor

2

2\sqrt{16/400} = \dfrac25

\dfrac15

(no

2

) and

5

(squared reciprocal) are the neighbours on the list.

Drill 4 more on parallel pair (h² = ab): factor as a perfect square, or use 2√((g² − ac)/(a(a + b)))

Concept 3 of 3

Product of the Perpendicular Distances From a Point to the Two Lines

Intuition

Factorise the pair and multiply the two point-to-line distances; the

\sqrt{\cdot}

denominators often cancel. For a homogeneous pair there is a closed form:

\dfrac{|ax_1^2 + 2hx_1y_1 + by_1^2|}{\sqrt{(a - b)^2 + 4h^2}}

Definition

$2x^2 - 5xy + 2y^2 = (2x - y)(x - 2y)$ ; from $(2, -1)$ : $P_1 = \dfrac{|4 + 1|}{\sqrt5} = \sqrt5$ , $P_2 = \dfrac{|2 + 2|}{\sqrt5} = \dfrac{4}{\sqrt5}$ ; product $4$ .
Closed form check: $\dfrac{|8 + 10 + 2|}{\sqrt{0 + 25}} = \dfrac{20}{5} = 4$ .
The closed form's denominator is $\sqrt{(a - b)^2 + 4h^2}$ , the same square root that appears in the angle formula's numerator squared plus $(a + b)^2$ .

Product of distances

P_1 P_2 = \frac{|ax_1^2 + 2hx_1y_1 + by_1^2|}{\sqrt{(a - b)^2 + 4h^2}}

Worked example

Find the product of the perpendicular distances from

(1, 2)

to the lines

x^2 - y^2 = 0

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3Pair of Straight LinesMODERATE

P_1

and

P_2

are perpendicular distances (in units) from point

(2,-1)

to the pair of lines

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

, then the value of

P_1 P_2

[Q145 · 16th May Shift 2 · 2023]

Adding the distances

The stem asks for the PRODUCT

P_1 P_2

. The sum

\sqrt5 + \dfrac{4}{\sqrt5} = \dfrac{9}{\sqrt5}

is not offered, but the squared product

16

-style and

5

10

are.

Drill 1 more on product of the perpendicular distances from a point to the two lines

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)

Condition for a Pair: abc + 2fgh − af² − bg² − ch² = 0
Pair condition
$\Delta = \begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix} = abc + 2fgh - af^2 - bg^2 - ch^2 = 0$
Parallel Pair (h² = ab): Factor as a Perfect Square, or Use 2√((g² − ac)/(a(a + b)))
Parallel pair
$h^2 = ab:\quad d = 2\sqrt{\frac{g^2 - ac}{a(a + b)}} = 2\sqrt{\frac{f^2 - bc}{b(a + b)}}$
Product of the Perpendicular Distances From a Point to the Two Lines
Product of distances
$P_1 P_2 = \frac{|ax_1^2 + 2hx_1y_1 + by_1^2|}{\sqrt{(a - b)^2 + 4h^2}}$

Watch out for (3)

Counting k = 0 as a pair
→ Condition for a Pair: abc + 2fgh − af² − bg² − ch² = 0
Reporting the gap squared, or halving it
→ Parallel Pair (h² = ab): Factor as a Perfect Square, or Use 2√((g² − ac)/(a(a + b)))
Adding the distances
→ Product of the Perpendicular Distances From a Point to the Two Lines

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