MHT-CET Maths · Straight Line

Forms of a Line, Intersections and Concurrency

Five ways to write a line — point-slope, two-point, intercept x/a + y/b = 1, normal x cos α + y sin α = p, general — plus the point where two lines meet and the determinant test for three lines through one point.

All Straight Line notes MHT-CET Maths strategy

Why this matters

30 PYQs at 13% HARD, the chapter's largest page. The intercept form carries the most stems (a line through a point with intercepts in a ratio, a triangle of given area with the axes, 1/a² + 1/b² = 1/p²); the normal form appears with the angle of the perpendicular given; medians and parallels test point-slope; and the concurrency determinant was set as a cubic in k in 2024. Three of its four HARD questions are the area-with-axes count and the concurrency cubic, both of which are careful casework, not new ideas.

Concept 1 of 5: Intercept Form x/a + y/b = 1: Ratios of Intercepts, Triangle Area and 1/a² + 1/b² = 1/p²

A line meeting the axes at

(a, 0)

and

(0, b)

\dfrac{x}{a} + \dfrac{y}{b} = 1

. It makes a right triangle of area

\dfrac{|ab|}{2}

with the axes, and its distance from the origin

p

satisfies

\dfrac{1}{p^2} = \dfrac{1}{a^2} + \dfrac{1}{b^2}

Definition

$b = 3a$ , through $(1, 3)$ : $\dfrac{1}{a} + \dfrac{3}{3a} = 1 \Rightarrow a = 2$ , $b = 6$ : $3x + y = 6$ .
Through $(2, 3)$ with area $12$ : $\dfrac{2}{a} + \dfrac{3}{b} = 1$ , $|ab| = 24$ . With $ab = 24$ : $2b + 3a = 24 \Rightarrow 3a^2 - 24a + 48 = 0 \Rightarrow a = 4$ , $b = 6$ (one line, $3x + 2y = 12$ ). With $ab = -24$ : $3a^2 + 24a + 48 = 0$ … gives $2b + 3a = -24$ , $3a^2 + 24a - 72 = 0$ , two real roots — two more lines. Three lines in all.
$P(-4, 1)$ divides $A(a, 0)$ , $B(0, b)$ in $1 : 2$ : $\left(\dfrac{2a}{3}, \dfrac{b}{3}\right) = (-4, 1) \Rightarrow a = -6$ , $b = 3$ : $x - 2y + 6 = 0$ .
Distance from the origin: $p = \dfrac{1}{\sqrt{1/a^2 + 1/b^2}}$ , hence $\dfrac{1}{a^2} + \dfrac{1}{b^2} = \dfrac{1}{p^2}$ .

Intercept form

\frac{x}{a} + \frac{y}{b} = 1,\qquad \text{area with axes} = \frac{|ab|}{2},\qquad \frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{p^2}

Worked example

A line through

(3, 4)

has

y

-intercept twice its

x

-intercept. Find the line.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 11th May Shift 1 · Q145Easy

Example 1 · Straight Line · Forms of a Line, Intersections and Concurrency

a and b are the intercepts made by a line on the coordinate axes. If

3a = b

and the line passes through

(1,3)

, then the equation of the line is

Forgetting the negative-product case

Area

12

means

|ab| = 24

, so

ab = -24

must be solved too; it supplies two of the three lines through

(2, 3)

. 'One' and 'two' are the options for the student who stopped at

ab = 24

Drill 4 more on intercept form x/a + y/b = 1: ratios of intercepts, triangle area and 1/a² + 1/b² = 1/p²

Concept 2 of 5: Normal Form x cos α + y sin α = p: the Perpendicular's Length and Direction

If the perpendicular from the origin to a line has length

p

and makes angle

\alpha

with the positive

x

-axis, the line is

x\cos\alpha + y\sin\alpha = p

. The direction

\alpha

fixes the signs;

p > 0

always.

Definition

$p = 2\sqrt2$ , $\alpha = 135^\circ$ : $-\dfrac{x}{\sqrt2} + \dfrac{y}{\sqrt2} = 2\sqrt2 \Rightarrow x - y + 4 = 0$ . The foot of the perpendicular is $(-2, 2)$ , in the second quadrant as $135^\circ$ demands; $x + y = 4$ has the right distance but the wrong direction ( $45^\circ$ ).
$p = 7$ , $\alpha = 120^\circ$ : $-\dfrac{x}{2} + \dfrac{\sqrt3 y}{2} = 7 \Rightarrow x - \sqrt3 y + 14 = 0$ .
To convert $ax + by + c = 0$ : divide by $\pm\sqrt{a^2 + b^2}$ so that the constant on the right is positive.
The line's own slope is $-\cot\alpha$ ; its inclination is $\alpha + 90^\circ$ .

Normal form

x\cos\alpha + y\sin\alpha = p \quad (p > 0),\qquad \text{foot of the perpendicular } (p\cos\alpha, p\sin\alpha)

Worked example

Write the line whose perpendicular from the origin has length

4

and makes

60^\circ

with the positive

x

-axis.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 10th May Shift 2 · Q139Moderate

Example 2 · Straight Line · Forms of a Line, Intersections and Concurrency

If the length of the perpendicular to a line from the origin is

2\sqrt{2}

units, which makes an angle of

135^\circ

with the X-axis, then the equation of line is

Matching the distance and ignoring the direction

Both

x + y = 4

and

x - y + 4 = 0

are

2\sqrt2

from the origin. Only one has its perpendicular at

135^\circ

; check the foot's quadrant.

Drill 1 more on normal form x cos α + y sin α = p: the perpendicular's length and direction

Concept 3 of 5: Point-Slope and Two-Point Forms: Medians, Parallels Through a Point, and Reading Intercepts Back

Given a point and a slope,

y - y_1 = m(x - x_1)

; given two points, the slope comes first. A median joins a vertex to the midpoint of the opposite side; a parallel through a point copies the slope. Once written, set

y = 0

and

x = 0

to read the intercepts.

Definition

$P(2, 2)$ , $Q(6, -1)$ , $R(7, 3)$ : midpoint $S\left(\tfrac{13}{2}, 1\right)$ , slope of $PS$ is $-\tfrac29$ . The parallel through $(1, -1)$ : $2x + 9y + 7 = 0$ ; intercepts $-\tfrac72$ and $-\tfrac79$ .
Median from $A(3, k)$ to $BC$ with $B(2, 1)$ , $C(-4, 5)$ has equation $x + 4y = p$ : the midpoint $(-1, 3)$ gives $p = 11$ ; then $3 + 4k = 11$ , $k = 2$ .
Two-point form: $\dfrac{y - y_1}{y_2 - y_1} = \dfrac{x - x_1}{x_2 - x_1}$ .
An intercept can be negative; the sign is part of the answer.

Point-slope

y - y_1 = m(x - x_1),\qquad \text{midpoint } \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)

Worked example

Find the median from

A(1, 4)

in the triangle with

B(3, -2)

C(7, 0)

, and its

x

-intercept.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 13th May Shift 2 · Q149Moderate

Example 3 · Straight Line · Forms of a Line, Intersections and Concurrency

PS is the median of the triangle with vertices at

P(2,2), Q(6,-1)

and

R(7,3)

, then the intercepts on the co-ordinate axes of the line passing through point

(1,-1)

and parallel to PS are respectively

Using the vertex instead of the midpoint

A median goes to the MIDPOINT of the opposite side. Joining

P

Q

R

gives a side, and the resulting intercepts are on the list.

Drill 1 more on point-slope and two-point forms: medians, parallels through a point, and reading intercepts back

Concept 4 of 5: Intersection of Two Lines, and a Line Through It With a Given Property

Solve the two equations to get the point, then write the required line through it — equal intercepts means

x + y = c

, a given slope means point-slope. For a parametric line, substitute one equation into the other and read the condition off the resulting expression.

Definition

$3x - y = 5$ , $x + 3y = 1$ meet at $\left(\tfrac85, -\tfrac15\right)$ ; the equal-intercept line through it is $x + y = \tfrac75$ , i.e. $5x + 5y - 7 = 0$ .
$3x + 4y = 9$ with $y = mx + 1$ : $x = \dfrac{5}{3 + 4m}$ . For integer $x$ with integer $m$ , $3 + 4m$ divides $5$ : $3 + 4m \in \{\pm1, \pm5\}$ gives integer $m$ only for $-1$ and $-5$ : $m = -1, -2$ . Two values.
Family through the intersection of $L_1 = 0$ and $L_2 = 0$ : $L_1 + \lambda L_2 = 0$ — useful when the point itself is ugly.
'Equal intercepts' includes the sign: $x + y = c$ . 'Equal in magnitude' would also allow $x - y = c$ .

Family through an intersection

L_1 + \lambda L_2 = 0 \qquad \text{equal intercepts: } x + y = c

Worked example

Find the line through the intersection of

x + 2y = 5

and

3x - y = 1

that is parallel to

4x + 3y = 0

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 3rd May Shift 2 · Q101Moderate

Example 4 · Straight Line · Forms of a Line, Intersections and Concurrency

The equation of the line passing through the point of intersection of the lines

3x-y=5

and

x+3y=1

and making equal intercepts on the axes is

Counting a divisor that gives a fractional m

3 + 4m = 1

has

m = -\tfrac12

; it is a divisor of

5

but not an integer

m

. Only two of the four divisors survive.

Drill 1 more on intersection of two lines, and a line through it with a given property

Concept 5 of 5: Concurrency of Three Lines: the 3 × 3 Determinant Is Zero

Three lines

a_ix + b_iy + c_i = 0

pass through one point iff the determinant of their coefficients vanishes. It is the same test as collinearity of three points, and later as coplanarity and the scalar triple product — one determinant in four costumes.

Definition

$kx + 2y + 2 = 0$ , $2x + ky + 3 = 0$ , $3x + 3y + k = 0$ : $\begin{vmatrix} k & 2 & 2 \\ 2 & k & 3 \\ 3 & 3 & k \end{vmatrix} = k^3 - 19k + 30 = (k - 2)(k - 3)(k + 5) = 0$ . $\sum k_i = 0$ — visible at once from Vieta, since there is no $k^2$ term.
$ax + by = c$ , $bx + cy = a$ , $cx + ay = b$ concurrent: $\begin{vmatrix} a & b & -c \\ b & c & -a \\ c & a & -b \end{vmatrix} = 0 \Rightarrow a^3 + b^3 + c^3 - 3abc = 0 \Rightarrow a + b + c = 0$ (or $a = b = c$ ).
Alternative: intersect two lines and substitute into the third — quicker when the numbers are small, worse when a parameter is involved.
The determinant vanishes also when two of the lines are parallel and the third is anything; check for a genuine common point when $k$ makes two lines parallel.

Concurrency

\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = 0

Worked example

Find

\lambda

x + 2y = 3

2x - y = 1

and

3x + \lambda y = 5

are concurrent.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 12th May Shift 2 · Q112Hard

Example 5 · Straight Line · Forms of a Line, Intersections and Concurrency

k_i

are possible values of

k

for which lines

kx+2y+2=0,\; 2x+ky+3=0

and

3x+3y+k=0

are concurrent, then

\sum k_i

has the value

Expanding the determinant wrong and losing the k term

The cubic is

k^3 - 19k + 30

, not

k^3 - 10k + 30

. Whatever the expansion, the answer to 'sum of the

k_i

' is the negative of the

k^2

coefficient —

0

here — so Vieta is the check.

Drill 1 more on concurrency of three lines: the 3 × 3 determinant is zero

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 (5)

Intercept Form x/a + y/b = 1: Ratios of Intercepts, Triangle Area and 1/a² + 1/b² = 1/p²
Intercept form
$\frac{x}{a} + \frac{y}{b} = 1,\qquad \text{area with axes} = \frac{|ab|}{2},\qquad \frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{p^2}$
Normal Form x cos α + y sin α = p: the Perpendicular's Length and Direction
Normal form
$x\cos\alpha + y\sin\alpha = p \quad (p > 0),\qquad \text{foot of the perpendicular } (p\cos\alpha, p\sin\alpha)$
Point-Slope and Two-Point Forms: Medians, Parallels Through a Point, and Reading Intercepts Back
Point-slope
$y - y_1 = m(x - x_1),\qquad \text{midpoint } \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$
Intersection of Two Lines, and a Line Through It With a Given Property
Family through an intersection
$L_1 + \lambda L_2 = 0 \qquad \text{equal intercepts: } x + y = c$
Concurrency of Three Lines: the 3 × 3 Determinant Is Zero
Concurrency
$\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = 0$

Watch out for (5)

Forgetting the negative-product case
→ Intercept Form x/a + y/b = 1: Ratios of Intercepts, Triangle Area and 1/a² + 1/b² = 1/p²
Matching the distance and ignoring the direction
→ Normal Form x cos α + y sin α = p: the Perpendicular's Length and Direction
Using the vertex instead of the midpoint
→ Point-Slope and Two-Point Forms: Medians, Parallels Through a Point, and Reading Intercepts Back
Counting a divisor that gives a fractional m
→ Intersection of Two Lines, and a Line Through It With a Given Property
Expanding the determinant wrong and losing the k term
→ Concurrency of Three Lines: the 3 × 3 Determinant Is Zero

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