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

19 PYQs at 47% 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; one stem is 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

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.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 20 April Shift II · Q114Moderate

Example 1 · Pair of Straight Lines · General Second-Degree Equation — Condition for a Pair, Parallel Lines and Distances

If the equation

kxy+ 10x+ 8y+ 16 = 0

represents a pair of lines, then

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)))

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

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 20 April Shift I · Q137Moderate

Example 2 · Pair of Straight Lines · General Second-Degree Equation — Condition for a Pair, Parallel Lines and Distances

The distance between the lines represented by the equation

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

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

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

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 16th May Shift 2 · Q145Moderate

Example 3 · Pair of Straight Lines · General Second-Degree Equation — Condition for a Pair, Parallel Lines and Distances

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

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.

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

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 19 questions

Related notes

Straight Line — distance between parallel lines and from a point →