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MHT-CET Maths · Formula sheet

Straight Line formulas

16 formulas and 16 common traps for MHT-CET Maths Straight Line, grouped by subtopic.

Full notes with worked examples

Slope, Angle Between Lines and Rotation

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Slope and Inclination: Parallel Means m₁ = m₂, Perpendicular Means m₁m₂ = −1

Slope

m=tan⁡θ=y2−y1x2−x1=−abparallel: m1=m2perpendicular: m1m2=−1m = \tan\theta = \frac{y_2 - y_1}{x_2 - x_1} = -\frac{a}{b} \qquad \text{parallel: } m_1 = m_2 \qquad \text{perpendicular: } m_1 m_2 = -1

Angle Between Two Lines: tan θ = |(m₁ − m₂)/(1 + m₁m₂)|

Angle between lines

tan⁡θ=∣m1−m21+m1m2∣one line vertical: tan⁡θ=∣1m∣\tan\theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right| \qquad \text{one line vertical: } \tan\theta = \left|\frac{1}{m}\right|

Lines Through a Point at a Given Angle: Solve the Angle Formula for m

Two slopes

∣m−m11+mm1∣=tan⁡α ⇒ m=m1±tan⁡α1∓m1tan⁡α(inclinations θ1±α)\left|\frac{m - m_1}{1 + m m_1}\right| = \tan\alpha \ \Rightarrow\ m = \frac{m_1 \pm \tan\alpha}{1 \mp m_1\tan\alpha} \quad(\text{inclinations } \theta_1 \pm \alpha)

Rotating a Line About a Point: Add or Subtract the Angle From the Inclination

Rotation

θnew=θ±α (+ anticlockwise),y−y0=tan⁡θnew (x−x0)\theta_{\text{new}} = \theta \pm \alpha \ (+\text{ anticlockwise}),\qquad y - y_0 = \tan\theta_{\text{new}}\,(x - x_0)

Bisector of an Angle at a Vertex, and the Reflection of a Line

Bisector direction

θbisector=θQP+θQR2reflected slope: the other root of ∣m−mL1+mmL∣=tan⁡θ\theta_{\text{bisector}} = \frac{\theta_{QP} + \theta_{QR}}{2} \qquad \text{reflected slope: the other root of } \left|\frac{m - m_L}{1 + m m_L}\right| = \tan\theta

Common traps

Reading a negative slope as a negative angle

Inclination lives in [0,π)[0, \pi). Slope −1-1 is 3π4\dfrac{3\pi}{4}; option π4\dfrac{\pi}{4} is the sign-blind answer.

Dropping the modulus and reporting the obtuse angle

m1−m21+m1m2\dfrac{m_1 - m_2}{1 + m_1 m_2} can come out negative; the acute angle uses its absolute value. tan⁡−1(−3)\tan^{-1}(-3) is not an option, but tan⁡−113\tan^{-1}\frac13 — the reciprocal slip — is.

Keeping only one root

The angle condition always yields TWO lines. Option lists pair the right line with a wrong partner (a sign flipped in the second equation); check both.

Rotating the wrong way

Clockwise subtracts. 30∘−15∘=15∘30^\circ - 15^\circ = 15^\circ gives slope 2−32 - \sqrt3; adding gives 45∘45^\circ and a line that is on the list as a distractor.

Bisecting the inclinations instead of the directions

QPQP has inclination 0∘0^\circ as a line but direction 180∘180^\circ from QQ. Averaging 0∘0^\circ and 60∘60^\circ gives 30∘30^\circ — the bisector of the OTHER angle at QQ, and its equation is offered.

Forms of a Line, Intersections and Concurrency

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

Intercept form

xa+yb=1,area with axes=∣ab∣2,1a2+1b2=1p2\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

xcos⁡α+ysin⁡α=p(p>0),foot of the perpendicular (pcos⁡α,psin⁡α)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−y1=m(x−x1),midpoint (x1+x22,y1+y22)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

L1+λL2=0equal intercepts: x+y=cL_1 + \lambda L_2 = 0 \qquad \text{equal intercepts: } x + y = c

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

Concurrency

∣a1b1c1a2b2c2a3b3c3∣=0\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = 0

Common traps

Forgetting the negative-product case

Area 1212 means ∣ab∣=24|ab| = 24, so ab=−24ab = -24 must be solved too; it supplies two of the three lines through (2,3)(2, 3). 'One' and 'two' are the options for the student who stopped at ab=24ab = 24.

Matching the distance and ignoring the direction

Both x+y=4x + y = 4 and x−y+4=0x - y + 4 = 0 are 222\sqrt2 from the origin. Only one has its perpendicular at 135∘135^\circ; check the foot's quadrant.

Using the vertex instead of the midpoint

A median goes to the MIDPOINT of the opposite side. Joining PP to QQ or RR gives a side, and the resulting intercepts are on the list.

Counting a divisor that gives a fractional m

3+4m=13 + 4m = 1 has m=−12m = -\tfrac12; it is a divisor of 55 but not an integer mm. Only two of the four divisors survive.

Expanding the determinant wrong and losing the k term

The cubic is k3−19k+30k^3 - 19k + 30, not k3−10k+30k^3 - 10k + 30. Whatever the expansion, the answer to 'sum of the kik_i' is the negative of the k2k^2 coefficient — 00 here — so Vieta is the check.

Section Formula, Midpoints and Rectangles

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Section Formula: Internal and External Division

Section formula

P=(mx2+nx1m+n,my2+ny1m+n)(internal);AOOB=d1d2 between parallel linesP = \left(\frac{m x_2 + n x_1}{m + n}, \frac{m y_2 + n y_1}{m + n}\right) \quad(\text{internal});\qquad \frac{AO}{OB} = \frac{d_1}{d_2} \text{ between parallel lines}

Midpoints Everywhere: Rectangle Centres, Missing Vertices and the Circumcentre

Rectangle centre

centre=midpoint of a diagonal;right angle at a vertex: mVA mVC=−1\text{centre} = \text{midpoint of a diagonal};\qquad \text{right angle at a vertex: } m_{VA}\, m_{VC} = -1

Common traps

Swapping the weights

In m:nm : n the weight mm goes on the SECOND point. (3⋅1+2⋅25,… )\left(\dfrac{3 \cdot 1 + 2 \cdot 2}{5}, \dots\right) gives (75,115)\left(\dfrac75, \dfrac{11}{5}\right) and a different kk.

Solving for the vertices without fixing c first

The line y=2x+cy = 2x + c has an unknown; the midpoint of the known diagonal lies on it and gives c=−4c = -4 in one line. Without that step the right-angle condition has two unknowns.

Distance — From a Point, Between Parallels, Along a Direction and the Foot of the Perpendicular

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Distance From a Point to a Line

Point to line

d=∣ax0+by0+c∣a2+b2d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}

Distance Between Parallel Lines, and the Square Between Them

Parallel lines

d=∣c1−c2∣a2+b2for ax+by+c1=0, ax+by+c2=0d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}} \quad\text{for } ax + by + c_1 = 0,\ ax + by + c_2 = 0

Foot of the Perpendicular: Intersect the Perpendicular Through the Point With the Line

Foot of the perpendicular

h−x1a=k−y1b=−ax1+by1+ca2+b2\frac{h - x_1}{a} = \frac{k - y_1}{b} = -\frac{ax_1 + by_1 + c}{a^2 + b^2}

Distance From a Point to a Line Measured Parallel to Another Line

Along a direction

dθ=∣ax1+by1+c∣∣acos⁡θ+bsin⁡θ∣d_\theta = \frac{|ax_1 + by_1 + c|}{|a\cos\theta + b\sin\theta|}

Common traps

Comparing |c| without dividing by √(a² + b²)

5x−2y=35x - 2y = 3 has the smallest constant AND the smallest distance, but 3x−4y+4=03x - 4y + 4 = 0 has ∣c∣=4>3|c| = 4 > 3 and distance 0.80.8, less than 513\dfrac{5}{\sqrt{13}} from the line with ∣c∣=5|c| = 5. Always divide.

Subtracting constants of differently scaled lines

4x+3y=204x + 3y = 20 and 8x+6y=308x + 6y = 30 are not 1010 apart; rescale to 4x+3y=154x + 3y = 15 first, giving 11.

Sign slips in the ratio form

The common ratio is MINUS ax1+by1+ca2+b2\dfrac{ax_1 + by_1 + c}{a^2 + b^2}. Dropping the minus sends the foot to the wrong side, (−5,4)(-5, 4)-style, which is never on the line — check the foot satisfies the line's equation.

Reporting the perpendicular distance

32\dfrac{3}{\sqrt2} is the perpendicular distance from (1,2)(1, 2) to x+y=0x + y = 0; measured along 3x−y=23x - y = 2 it is 3104\dfrac{3\sqrt{10}}{4}. 'Measured parallel to' changes the answer.

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