NDA Coordinates and Straight Lines previous year questions with solutions

Answer: (a) on a straight line

Method. AP in both coordinates makes Y the mid-point of XZ

a, b, c in AP: b = a + c2; p, q, r in AP: q = p + r2.
So Y(b, q) is the mid-point of XZ.
Hence X, Y, Z lie on a straight line.

Answer: (a) y-axis

Method. equal x-coordinates give a vertical line

a, b, c in AP with b = c: a = 2b − c = b, so a = b = c.
All three points have the same x-coordinate a: they lie on the line x = a.
x = a is parallel to the y-axis.

Answer: (a) (1, ⁠√2⁠ − 1)

Method. incentre = (aA + bB + cC)/⁠(a + b + c)

Side lengths: a = BC = 2, b = CA = ⁠√2,⁠ c = AB = ⁠√2.⁠
x = 2(1) + ⁠√2(0)⁠ + ⁠√2(2)2⁠ + 2⁠√2⁠ = 2 + 2⁠√22⁠ + 2⁠√2⁠ = 1.
y = 2(1) + ⁠√2(0)⁠ + ⁠√2(0)2⁠ + 2⁠√2⁠ = 11 + ⁠√2⁠ = ⁠√2⁠ − 1.
P = (1, ⁠√2⁠ − 1).

Shortcut. Isosceles with AB = AC, so P lies on x = 1; inradius r = Areas = 11 + ⁠√2⁠ = ⁠√2⁠ − 1.

Answer: (a) (1 + 2⁠√5,⁠ 5 + 4⁠√5),⁠ (1 − 2⁠√5,⁠ 5 − 4⁠√5)⁠

Method. parametric (distance) form of a line through a point on it

R(1, 5) lies on y = 2x + 3, since 2(1) + 3 = 5.
The line has slope 2, so a unit vector along it is (1⁠√5,⁠ 2⁠√5).⁠
Points at distance 2 from R along the line: (1 ± 2⁠√5,⁠ 5 ± 4⁠√5)⁠ with the same sign taken.
P, Q = (1 + 2⁠√5,⁠ 5 + 4⁠√5)⁠ and (1 − 2⁠√5,⁠ 5 − 4⁠√5).⁠

Answer: (b) (73, 73)

Method. mid-point formula to recover vertices, then centroid

Mid-point of AB is (-1, 2), so B = 2(-1, 2) − (3, 5) = (-5, -1).
Mid-point of AC is (6, 4), so C = 2(6, 4) − (3, 5) = (9, 3).
Centroid = (3 − 5 + 93, 5 − 1 + 33) = (73, 73).

Shortcut. Since B = 2M_AB - A and C = 2M_AC - A, centroid = 2M_AB + 2M_AC - A3 = (-2 + 12 − 33, 4 + 8 − 53).

Answer: (d) collinear

Method. every point is a scalar multiple of (a, b)

(-a, -b) = -1·(a, b), (0, 0) = 0·(a, b), (a, b) = 1·(a, b), (a², ab) = a·(a, b).
All four lie on the line through the origin in direction (a, b), i.e. bx − ay = 0.
So the points are collinear.

Answer: (b) 1 : 2

Method. ratio in which a line divides a segment: -L(P): L(Q)

L(x, y) = x + y − 4. L(P) = -1 + 1 − 4 = -4; L(Q) = 5 + 7 − 4 = 8.
PR: RQ = -L(P): L(Q) = 4 : 8 = 1 : 2.
Check: R = (2(-1) + 53, 2(1) + 73) = (1, 3), and 1 + 3 = 4.

Answer: (c) 1

Method. collinearity by equal slopes

Slope between (5p², 10p) and (5q², 10q) = 10(p − q)5(p² - q²) = 2p + q (p ≠ q).
Slope between (-5, 0) and (5p², 10p) = 10p5p² + 5 = 2pp² + 1.
Equal: p² + 1 = p(p + q) = p² + pq, so pq = 1.

Answer: (a) 1 only

Method. test each point against one line; quadrant check by counter-example

(p, p − 3): y = x − 3. (q + 3, q): y = x − 3. (6, 3): 3 = 6 − 3. All three lie on y = x − 3, so statement 1 is correct.
Statement 2: p = 0 gives (0, -3), which is not in the first quadrant; so statement 2 is wrong.
Answer: 1 only.

Answer: (c) ad = bc

Method. collinearity: area determinant equals zero

Collinear ⇔  a b 1 c d 1 a − c b − d 1  = 0.
Expanding: a(d − b + d) − b(c − a + c) + c(b − d) − d(a − c)
= 2ad − ab − 2bc + ab + bc − cd − ad + cd = ad − bc.
So the points are collinear when ad = bc.

Shortcut. Slope from (a, b) to (c, d) is d − bc − a; from (c, d) to (a − c, b − d) is b − 2da − 2c; equating and cross-multiplying gives ad = bc.

Answer: (b) 10

Method. mid-point theorem: BC = 2 DE

DE joins the mid-points of AB and AC, so DE ∥ BC and DE = 12BC.
DE = ⁠√((5⁠ −⁠ 2)² + (9 −⁠ 5)²) = ⁠√(9⁠ + 16) = 5.
BC = 2 × 5 = 10.

Answer: (a) |2m sin(α − β)|

Method. distance formula and 1 − cos 2θ = 2 sin² θ

PQ² = m²[(cos 2α − cos 2β)² + (sin 2α − sin 2β)²] = m²[2 − 2cos(2α − 2β)].
2 − 2cos(2α − 2β) = 4 sin²(α − β).
PQ = |2m sin(α − β)|.

Options (c) and (d) are printed identically, but both are wrong, so the answer (a) is unaffected.

Answer: (d) Statement I is false, but Statement II is true

Method. cos α from the dot product of OP and OQ; cosine rule

With O the origin, cos α = OP·OQ|OP||OQ| = mr + ns⁠√((m²⁠ + n²)(r² + s²)).
The expression ms − nr⁠√(…)⁠ is sin α (up to sign), not cos α; e.g. P(1, 0), Q(0, 1) give α = 90°, cos α = 0, but ms − nr = 1. Statement I is false.
Statement II is the cosine rule: true.
Statement I is false, Statement II is true.

Answer: (d) |(x₁ - x₂)(x₂ - x₃)(x₃ - x₁)2x₁x₂x₃|

Method. area = 12|determinant|, then clear the fractions

Area = 12| x₁ 1/⁠x₁ 1; x₂ 1/⁠x₂ 1; x₃ 1/⁠x₃ 1 |.
Multiply row i by x_i: the determinant = 1x₁x₂x₃| x₁² 1 x₁; x₂² 1 x₂; x₃² 1 x₃ |.
A cyclic shift of columns turns this into the Vandermonde determinant | 1 x₁ x₁²; 1 x₂ x₂²; 1 x₃ x₃² | = (x₂ - x₁)(x₃ - x₁)(x₃ - x₂).
In absolute value this is |(x₁ - x₂)(x₂ - x₃)(x₃ - x₁)|.
Area = |(x₁ - x₂)(x₂ - x₃)(x₃ - x₁)2x₁x₂x₃|.

Answer: (a) 12⁠√175⁠

Method. internal and external section formulae

Internal point (2 : 3): (2.5 + 3.45, 2.7 + 3.35) = (225, 235).
External point (2 : 3): (2.5 − 3.42 − 3, 2.7 − 3.32 − 3) = (2, -5).
Differences: 225 − 2 = 125, 235 + 5 = 485.
Distance = 125⁠√(1⁠ + 16) = 12⁠√175.⁠

Answer: (d) collinear

Method. every point is a scalar multiple of (a, b)

(0, 0) = 0·(a, b), (-a, -b) = -1·(a, b) and (ab, b²) = b·(a, b).
So all four points lie on the line through the origin in the direction (a, b), i.e. bx − ay = 0.
Check (ab, b²): b·ab − a·b² = 0. (holds)
The points are collinear.

Answer: (d) (1, 73)

Method. midpoint formula to recover the vertices, then the centroid

Let A = (1, 1). The midpoint of AB is (-1, 2), so B = (2(-1) − 1, 2(2) − 1) = (-3, 3).
The midpoint of AC is (3, 2), so C = (2(3) − 1, 2(2) − 1) = (5, 3).
Centroid = (1 − 3 + 53, 1 + 3 + 33) = (1, 73).

Answer: (d) (1, 1⁠√3)⁠

Method. the triangle is equilateral, so the incentre is the centroid

AB = ⁠√(1⁠ + 3) = 2, BC = 2, CA = ⁠√(1⁠ + 3) = 2: the triangle is equilateral.
In an equilateral triangle the incentre coincides with the centroid.
Incentre = (1 + 0 + 23, ⁠√3⁠ + 0 + 03) = (1, ⁠√33)⁠ = (1, 1⁠√3).⁠

Answer: (b) 3 : 4

Method. section formula with ratio k: 1

Let C divide AB in the ratio k: 1. Then x = 2k − 2k + 1 = -27.
14k − 14 = -2k − 2, so 16k = 12 and k = 34.
Check y: -4k − 2k + 1 = -3 − 27/⁠4 = -207. (holds)
Ratio = 3 : 4.

Answer: (a) 5, 2

Method. centroid formula, coordinate by coordinate

x-coordinate: 7 + y + 93 = 6, so y = 2.
y-coordinate: x − 6 + 103 = 3, so x = 5.
x, y = 5, 2.

Answer: (c) 60°

Method. angle between two lines from slopes

Slope of L: -1⁠√3;⁠ slope of x - ⁠√3y⁠ = 0: 1⁠√3.⁠
tan α = |1⁠√3⁠ + 1⁠√31⁠ − 13| = 2/⁠√32/⁠3⁠ = ⁠√3.⁠
α = 60°.

Answer: (c) 60°

Method. inclination of L, then angle with the y-axis

Slope of L = -1⁠√3,⁠ so L makes 150° with the positive x-axis.
The positive y-axis is at 90°, so the angle between L and the positive y-axis is 150° - 90° = 60°.

Answer: (d) 1⁠√2⁠

Method. angle between lines from their normal vectors

Line 1: px + qy = p + q, normal (p, q).
Line 2: (p + q)x + (q − p)y = 2q, normal (p + q, q − p).
Dot product: p(p + q) + q(q − p) = p² + q².
|(p, q)| = ⁠√(p²⁠ + q²); |(p + q, q − p)| = ⁠√(2(p²⁠ + q²)).
cos θ = p² + q²⁠√2⁠ (p² + q²) = 1⁠√2,⁠ so θ = 45° and sin θ = 1⁠√2.⁠

Answer: (a) mn − 1 = 0

Method. a₁a₂ + b₁b₂ = 0 for perpendicular lines

For a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 to be perpendicular: a₁a₂ + b₁b₂ = 0.
m²·n² + n·(-m) = 0 ⟹ mn(mn − 1) = 0.
For non-zero m, n: mn − 1 = 0.

Answer: (d) Infinitely many

Method. x = 5 is a straight line in the xy-plane

In the xy-plane, x = 5 places no condition on y, so every point (5, y) satisfies it.
It is the vertical line through (5, 0), which contains infinitely many points.

Answer: (b) 12

Method. angle between two lines, then base angle = 90° - A/⁠2

Slopes: AB has m₁ = 7, AC has m₂ = -1.
tan A = |7 − (-1)1 + 7(-1)| = 86 = 43, taking the acute angle since the triangle is acute angled.
Equal angles: θ = 180° - A2 = 90° - A2, so cotθ = tanA2.
With t = tanA2: 2t1 − t² = 43 ⟹ 2t² + 3t − 2 = 0 ⟹ t = 12 (positive root).
cotθ = 12.

Answer: (b) -1, 2

Method. parallel lines: equal slopes (and not the same line)

Slope of the given line = -k − 3-(5 − k²) = k − 35 − k²; slope of x + y = 1 is -1.
k − 35 − k² = -1 gives k − 3 = k² - 5, i.e. k² - k − 2 = 0, (k − 2)(k + 1) = 0.
k = 2: -x − y − 4 = 0, i.e. x + y + 4 = 0; k = -1: -4x − 4y + 14 = 0, i.e. x + y = 72. Both are parallel to (not the same as) x + y = 1.
k = -1, 2.

Answer: (c) a² - b² = 0

Method. parallel lines have equal slopes

Slopes are -ab and -ba.
Equal: a² = b², i.e. a² - b² = 0.
(When a = b the two equations are the same line, a coincident case of equal slopes; a = -b gives distinct parallel lines when c ≠ 0. The slope condition covering both is a² - b² = 0.)

Answer: (c) pq − 1 = 0

Method. perpendicular lines: product of slopes is -1

y + px = 1 has slope -p; y − qx = 2 has slope q.
Perpendicular: (-p)(q) = -1, so pq = 1, i.e. pq − 1 = 0.

Answer: (b) 30°

Method. angle between lines from their inclinations

x − 2 = 0 is vertical: inclination 90°.
⁠√3x⁠ − y − 2 = 0 has slope ⁠√3:⁠ inclination 60°.
Acute angle between them = 90° - 60° = 30°.

Answer: (b) 120°

Method. slopes as tangents of standard angles

tan 15° = 2 - ⁠√3⁠ and tan 75° = 2 + ⁠√3.⁠
The lines are inclined at 15° and 75°, so the angle between them is 75° - 15° = 60°.
The obtuse angle is 180° - 60° = 120°.

Shortcut. tan θ = |m₁ - m₂1 + m₁m₂| = 2⁠√31⁠ + 1 = ⁠√3,⁠ so θ = 60° and the obtuse angle is 120°.

Answer: (d) only when at least one of a and b is non-zero

Method. a first-degree equation is a line only if it really contains x or y

If a = b = 0 the equation reads c = 0: no points (c ≠ 0) or the whole plane (c = 0), never a line.
If at least one of a, b is non-zero, the equation is first degree and its graph is a straight line (vertical when b = 0, horizontal when a = 0).
So it represents a line exactly when at least one of a and b is non-zero.

Answer: (d) (π − 2β + 2α)2

Method. angle between the normals of the two lines

Normal to the first line: (cos α, sin α); normal to the second: (sin β, -cos β); both unit vectors.
cos θ = cos α sin β − sin α cos β = sin(β − α) = cos(π2 − β + α).
So θ = π2 − β + α = (π − 2β + 2α)2.

Answer: (c) |lm' - l'mll' + mm'|

Method. tan θ = |m₁ - m₂1 + m₁m₂| with m = -coefficient of xcoefficient of y

m₁ = -lm, m₂ = -l'm'.
m₁ - m₂ = l'm − lm'mm', 1 + m₁m₂ = mm' + ll'mm'.
θ = |l'm − lm'll' + mm'| = |lm' - l'mll' + mm'|.

Answer: (b) tan^-1 (4m²n²m⁴ - n⁴)

Method. tan θ = |m₁ - m₂1 + m₁m₂|

m₁ = n(m + n)m(m − n), m₂ = n(m −⁠ n)m(m + n).
m₁ - m₂ = nm·(m + n)² - (m − n)²m² - n² = 4n²m² - n².
1 + m₁m₂ = 1 + n²m² = m² + n²m².
tan θ = 4n²m² - n²·m²m² + n² = 4m²n²m⁴ - n⁴.
θ = tan^-1 (4m²n²m⁴ - n⁴).

Answer: (b) α > β

Method. inclinations of the lines

x + y − 3 = 0 has slope -1 and x − y + 3 = 0 has slope 1; m₁m₂ = -1, so α = 90°.
x - ⁠√3y⁠ + 2⁠√3⁠ = 0 has slope 1⁠√3⁠ (inclination 30°); ⁠√3x⁠ − y + 1 = 0 has slope ⁠√3⁠ (inclination 60°).
β = 60° - 30° = 30°.
α = 90° > β = 30° (α = 3β, not 2β).

Answer: (a) tan^-1 (12⁠√6)⁠

Method. tan θ = |m₁ - m₂1 + m₁m₂|

m₁ = -⁠√2√3,⁠ m₂ = -⁠√3√2,⁠ so m₁m₂ = 1.
m₁ - m₂ = ⁠√3√2⁠ − ⁠√2√3⁠ = 3 − 2⁠√6⁠ = 1⁠√6.⁠
tan θ = 1/⁠√61⁠ + 1 = 12⁠√6,⁠ so θ = tan^-1 (12⁠√6).⁠

Answer: (b) 1

Method. intercepts, mid-point, distance from origin

y = 0: x = sin 2θsin θ = 2 cos θ, so P = (2 cos θ, 0).
x = 0: y = sin 2θcos θ = 2 sin θ, so Q = (0, 2 sin θ).
M = (cos θ, sin θ), so OM = ⁠√(cos²θ⁠ + sin²θ) = 1.

Answer: (a) 3x − 5y − 15 = 0

Method. intercepts as roots of t² - 2t − 15 = 0, then intercept form

p and q are roots of t² - 2t − 15 = 0, i.e. (t − 5)(t + 3) = 0.
p > q gives p = 5, q = -3.
x5 + y-3 = 1 ⟹ 3x − 5y = 15, i.e. 3x − 5y − 15 = 0.

Answer: (b) x + y − 3 = 0

Method. section formula, then intercept form

Let P(a, 0), Q(0, b) with a, b > 0; PN: NQ = 1 : 2.
N = (2·a + 1.03, 2.0 + 1·b3) = (2a3, b3) = (2, 1).
a = 3, b = 3.
Line: x3 + y3 = 1, i.e. x + y − 3 = 0.

Answer: (a) 2

Method. intercepts by setting y = 0 and x = 0

y = 0: x = 2a²a² + b²; x = 0: y = 2b²a² + b².
Sum = 2(a² + b²)a² + b² = 2.

Answer: (c) ⁠√3√7⁠ x + 2⁠√7⁠ y = ⁠√7⁠

Method. normal (perpendicular) form: divide by ⁠√(a²⁠ + b²)

The perpendicular (normal) form is x cos α + y sin α = p with p > 0.
Divide ⁠√3x⁠ + 2y = 7 by ⁠√((√3)²⁠ + 2²) = ⁠√7.⁠
⁠√3√7⁠ x + 2⁠√7⁠ y = 7⁠√7⁠ = ⁠√7.⁠
(a) is slope-intercept form and (b) intercept form; (d) has the wrong right side.

Answer: (b) x + 3y = 10xy

Method. intercept form through a fixed point, then midpoint substitution

Solve x + 2y = 1 and 2x − y = 1: y = 2x − 1, so 5x − 2 = 1, giving the point (35, 15).
Let the line be xa + yb = 1 with A(a, 0), B(0, b); it passes through the point: 35a + 15b = 1.
Mid-point of AB is (h, k) = (a2, b2), so a = 2h, b = 2k.
310h + 110k = 1 ⟹ 3k + h = 10hk.
Locus: x + 3y = 10xy.

Answer: (b) x sinθ − y cosθ + 1 = 0

Method. line perpendicular to ax + by = c is bx − ay = k

A line perpendicular to x cosθ + y sinθ = 9 has the form x sinθ − y cosθ = k.
Through (-sinθ, cosθ): k = -sin²θ − cos²θ = -1.
Equation: x sinθ − y cosθ + 1 = 0.

Answer: (c) 8⁠√6⁠

Method. normal form x cos α + y sin α = p and its intercepts

Normal form: x cos 15° + y sin 15° = 4.
x-intercept = 4cos 15°, y-intercept = 4sin 15°.
Sum = 4(sin 15° + cos 15°)sin 15° cos 15°.
sin 15° cos 15° = 12 sin 30° = 14; (sin 15° + cos 15°)² = 1 + sin 30° = 32, so sin 15° + cos 15° = ⁠√62.⁠
Sum = 4 × ⁠√62⁠ × 4 = 8⁠√6.⁠

Answer: (c) Both 1 and 2

Method. slope tan 75° = 2 + ⁠√3⁠ = 12 - ⁠√3⁠

The line is y = x tan 75° = (2 + ⁠√3)x.⁠
12 - ⁠√3⁠ = 2 + ⁠√34⁠ − 3 = 2 + ⁠√3,⁠ so (1, 12 - ⁠√3)⁠ satisfies y = (2 + ⁠√3)x:⁠ statement 1 is correct.
A line through the origin with positive slope has x and y of the same sign at every other point, so it runs through the first and third quadrants only: statement 2 is correct.
Answer: both.

Answer: (b) 4x + 3y − 24 = 0

Method. mid-point of the intercept segment gives intercepts 2h and 2k

The segment between the axes has ends (a, 0) and (0, b) with mid-point (3, 4), so a = 6, b = 8.
Line: x6 + y8 = 1, i.e. 4x + 3y − 24 = 0.

Shortcut. Line with (h, k) as mid-point of its intercepts: xh + yk = 2, i.e. x3 + y4 = 2.

No correct option. This question is defective as printed in UPSC's paper.

Method. slope of AC from the coordinates of the equilateral triangle

A(0, -4), B(0, 4); C is on the x-axis at distance ⁠√32⁠ × 8 = 4⁠√3⁠ from the origin, but on either side.
C(4⁠√3,⁠ 0): slope of AC = 44⁠√3⁠ = 1⁠√3;⁠ line through (8, 0): y = 1⁠√3(x⁠ − 8), i.e. x - ⁠√3⁠ y − 8 = 0 (option a).
C(-4⁠√3,⁠ 0): slope of AC = -1⁠√3;⁠ line: x + ⁠√3⁠ y − 8 = 0 (option b).
The question does not fix the side of C, so two options are correct.

The data do not say on which side of the y-axis C lies: C(4⁠√3,⁠ 0) gives option (a) and C(-4⁠√3,⁠ 0) gives option (b), so two options are correct; fixing C on the positive x-axis would give (a).

Answer: (c) x + y + 1 = 0

Method. intercept form with a = b

Equal intercepts: xa + ya = 1, i.e. x + y = a.
Through (1, -2): a = 1 − 2 = -1.
Line: x + y + 1 = 0.

Answer: (a) x − y = 0

Method. locus of mid-point of intercepts

x + y = p meets the axes at (p, 0) and (0, p).
Mid-point (h, k) = (p2, p2), so h = k for every p.
Locus: x − y = 0.

Answer: (b) x + y = 5

Method. perpendicular slope is the negative reciprocal

y = x has slope 1, so the required slope is -1.
y − 2 = -(x − 3), i.e. x + y = 5.

Answer: (d) x - ⁠√3⁠ y − 2⁠√3⁠ = 0

Method. slope-intercept form y = mx + c

m = tan 30° = 1⁠√3;⁠ intercept 2 on the negative y-axis gives c = -2.
y = x⁠√3⁠ − 2; multiply by ⁠√3:⁠ ⁠√3y⁠ = x − 2⁠√3.⁠
x - ⁠√3y⁠ − 2⁠√3⁠ = 0.

Answer: (b) 2x + y = 7

Method. intercept form xa + y2a = 1

Let the x-intercept be a, so the y-intercept is 2a: xa + y2a = 1, i.e. 2x + y = 2a.
Through (2, 3): 4 + 3 = 2a, so 2a = 7 (a = 72 > 0, as required).
Line: 2x + y = 7.

Answer: (d) Either x + y = 5 or 9x + 4y = 30

Method. section formula on the intercepts A(a, 0), B(0, b), both orders of the ratio

If (2, 3) divides AB with AP: PB = 3 : 2: (2, 3) = (2a5, 3b5), so a = 5, b = 5: x + y = 5.
If it divides BA in 3 : 2 (AP: PB = 2 : 3): (2, 3) = (3a5, 2b5), so a = 103, b = 152.
Then 3x10 + 2y15 = 1; multiply by 30: 9x + 4y = 30.
Either x + y = 5 or 9x + 4y = 30.

Answer: (a) y + ⁠√3x⁠ + 5 = 0

Method. slope-intercept form y = mx + c

m = tan 120° = -⁠√3;⁠ intercept on the negative y-axis is 5, so c = -5.
y = -⁠√3x⁠ − 5, i.e. y + ⁠√3x⁠ + 5 = 0.

Answer: (a) 2x + 3y − 4 = 0

Method. a parallel line keeps the same x and y coefficients

Required line: 2x + 3y + k = 0.
Through (-1, 2): -2 + 6 + k = 0, so k = -4.
Line: 2x + 3y − 4 = 0.

Answer: (c) 3x − 6y − 5 = 0

Method. line midway between parallel lines: average the constants after matching coefficients

Make the coefficients equal: 2x − 4y − 7 = 0 becomes 6x − 12y − 21 = 0.
The other line is 6x − 12y + 1 = 0.
Midway line: 6x − 12y + -21 + 12 = 0, i.e. 6x − 12y − 10 = 0.
Divide by 2: 3x − 6y − 5 = 0.

Answer: (c) 13/⁠5 units

Method. reflect P in the x-axis; the image, N and Q are collinear

Image of P(1, 2) in the x-axis: P′(1, -2).
The reflected ray QN extended passes through P′, so N is where P'Q meets y = 0.
P'Q: slope 3 − (-2)5 − 1 = 54; y + 2 = 54(x − 1).
y = 0: x − 1 = 85, so N = (135, 0).
ON = 135 units.

Shortcut. N divides P'Q in the ratio 2 : 3 (the distances from the x-axis), so x = 3(1) + 2(5)5 = 135.

Answer: (a) y = 10k

Method. cot A and cot B from the coordinates of C

Take A(0, 0), B(10, 0) on the x-axis and C(x, y) with y > 0; let F(x, 0) be the foot of the perpendicular from C.
cot A = AFCF = xy, cot B = FBCF = 10 − xy.
1tan A + 1tan B = x + 10 − xy = 10y = 1k.
Locus: y = 10k (a line parallel to AB).

Answer: (b) concurrent lines

Method. family of lines through a fixed point

Rewrite: 4y − 2 = m(x − 1).
For every m, x = 1, y = 12 satisfies the equation.
So all the lines pass through (1, 12): they are concurrent.

Answer: (d) (c² + d² - a² - b²)/⁠2

Method. locus equidistant from two points (perpendicular bisector)

(x − a)² + (y − b)² = (x − c)² + (y − d)².
-2ax + a² - 2by + b² = -2cx + c² - 2dy + d².
2(a − c)x + 2(b − d)y + (c² + d² - a² - b²) = 0.
Divide by 2: (a − c)x + (b − d)y + c² + d² - a² - b²2 = 0, so k = (c² + d² - a² - b²)/⁠2.

Answer: (b) (3, 1)

Method. perpendicular bisector in parametric form

P = (3 + 12, -1 + 12) = (2, 0).
Slope of AB = 1 − (-1)1 − 3 = -1, so the perpendicular bisector has slope 1: y = x − 2.
Unit vector along it: (1⁠√2,⁠ 1⁠√2);⁠ at distance ⁠√2⁠ from P: Q = (2 ± 1, 0 ± 1).
Q = (3, 1) or (1, -1); of the options, (3, 1).

Answer: (b) (2, -72)

Method. factorise the condition as a difference of squares

16p² + 49q² - 56pq = 4r² gives (4p − 7q)² = (2r)², so 4p − 7q = ±2r, i.e. 2p − 72q = ±r.
At (2, -72): px + qy = 2p − 72q = ±r, so one of px + qy + r = 0, px + qy − r = 0 holds.
Hence (2, -72) lies on the pair of lines.

Answer: (a) 2

Method. line through the intersection point; intercepts a and -a for the fourth quadrant

Solve x + 2y + 2 = 0 and 2x − 3y − 3 = 0: x = -2 − 2y, so -4 − 4y − 3y = 3, y = -1, x = 0. The point is (0, -1).
Equal intercepts in the fourth quadrant: x-intercept a > 0, y-intercept -a, so xa − ya = 1, i.e. x − y = a.
Through (0, -1): a = 1. Intercepts 1 and -1; sum of absolute values = 2.

Answer: (c) 4x − 6y + 5a = 0

Method. equidistant point: equate squared distances

(x − 2a)² + y² = x² + (y − 3a)².
-4ax + 4a² = -6ay + 9a², so -4ax + 6ay − 5a² = 0.
Divide by -a (a > 0): 4x − 6y + 5a = 0.

Answer: (d) (1, -1)

Method. family of lines through a fixed point: A(…) + C(…) = 0

A, B, C in AP gives 2B = A + C.
The line becomes Ax + (A + C)y + C = 0, i.e. A(x + y) + C(y + 1) = 0.
This holds for all A, C when x + y = 0 and y + 1 = 0, i.e. at (1, -1).

Shortcut. Since A − 2B + C = 0, compare with Ax + 2By + C = 0: x = 1, y = -1 satisfies it for every AP.

Answer: (b) y = 2x

Method. the mirror is the perpendicular bisector of the point and its image

Mid-point of (-4, 2) and (4, -2) is (0, 0), so the mirror passes through the origin.
Slope of the joining segment = -2 − 24 + 4 = -12, so the mirror has slope 2.
Mirror: y = 2x.

Answer: (c) 5

Method. the segment to the foot of the perpendicular is along the normal of the line

Normal direction of 3x − 4y − 5 = 0 is (3, -4).
The segment from (0, k) to the foot (3, 1) is (3, 1 − k), and it must be parallel to (3, -4).
33 = 1 − k-4 gives 1 − k = -4, so k = 5.
Check: (3, 1) lies on the line, 9 − 4 − 5 = 0.

Shortcut. Slope of the perpendicular 1 − k3 must equal -43, so k = 5.

Answer: (d) Neither 1 nor 2

Method. perpendicular distance |aα + bβ + c|⁠√(a²⁠ + b²)

1. The line is x cos θ + y sin θ − p = 0 and cos² θ + sin² θ = 1, so the distance is |α cos θ + β sin θ − p|, not |α cos θ + β sin θ + p|. Incorrect.
2. xa + yb = 1 is bx + ay − ab = 0, so the distance is |bα + aβ − ab⁠√(a²⁠ + b²)|; the statement has a and b interchanged in the numerator. Incorrect.
Neither 1 nor 2.

Answer: (c) 6

Method. the third line passes through the intersection of the first two

Put y = x + 5 in 4x + 3y = 1: 4x + 3x + 15 = 1, so x = -2, y = 3.
5y + bx = 3 at (-2, 3): 15 − 2b = 3, so b = 6.

Answer: (d) 1, 2 and 3

Method. distance between parallel lines |c₁ - c₂|⁠√(a²⁠ + b²)

1. mx − y + c₁ = 0 and mx − y + c₂ = 0: distance |c₁ - c₂|⁠√(m²⁠ + 1). Correct.
2. The general formula itself. Correct.
3. x = c₁ and x = c₂ are vertical lines; the distance is the gap in x, |c₁ - c₂|. Correct.
1, 2 and 3.

Answer: (a) 20x + 25y − 54 = 0

Method. find the intersection, then keep the coefficients 4 and 5

The lines are 3x + 2y = 6 and 2x + 3y = 6; subtracting gives x = y, so 5x = 6 and the point is (65, 65).
Required line: 4x + 5y = k with k = 4·65 + 5·65 = 545.
Multiply by 5: 20x + 25y − 54 = 0.

Answer: (c) 5x − 5y + 18 = 0

Method. find the intersection, then use the parallel slope

From 2x − y + 5 = 0, y = 2x + 5; then x + 2(2x + 5) − 3 = 0 gives x = -75, y = 115.
y − x + 10 = 0 has slope 1.
Line: y − 115 = x + 75, i.e. y = x + 185.
Multiply by 5: 5x − 5y + 18 = 0.

Answer: (c) 1 and 2 only

Method. perpendicular distance from the origin, p = |c|⁠√(a²⁠ + b²)

1. ax + by − c = 0: p = |c|⁠√(a²⁠ + b²), so p² = c²a² + b². Correct.
2. xa + yb − 1 = 0: p = 1⁠√(1/⁠a²⁠ + 1/⁠b²), so 1p² = 1a² + 1b². Correct.
3. mx − y + c = 0: p = |c|⁠√(1⁠ + m²), so 1p² = 1 + m²c², not 1 + m² + c²c². Incorrect.
1 and 2 only.

Answer: (c) Equal to each other

Method. 2x + 11y = 5 is a bisector of the angle between the two lines

Distances from (x, y): |24x + 7y − 20|25 and |4x − 3y − 2|5.
The angle bisectors are 24x + 7y − 20 = ±5(4x − 3y − 2).
With the plus sign: 24x + 7y − 20 = 20x − 15y − 10, i.e. 4x + 22y − 10 = 0, which is 2x + 11y = 5.
So every point of 2x + 11y = 5 is equidistant from the two lines: the perpendiculars are equal.
Check with (0, 511): |35/⁠11 − 20|25 = 3755 and |-15/⁠11 − 2|5 = 3755. (holds)

Answer: (b) 3/⁠10

Method. distance between parallel lines |c₁ - c₂|⁠√(a²⁠ + b²)

Write the second line as 3x + 4y = 152.
Distance = |9 − 15/⁠2|⁠√(9⁠ + 16) = 3/⁠25 = 310.

Answer: (d) ⁠√172⁠ units

Method. move from the point along the given direction until the line is met

4x + y = 4 has slope -4, so a direction along it is (1, -4), of length ⁠√17.⁠
A point on that path: (1 + t, 3 − 4t). It meets 2x + 3y = 6 when 2 + 2t + 9 − 12t = 6, so t = 12.
Distance = |t|·⁠√17⁠ = ⁠√172⁠ units.

Answer: (b) 21x − 46y + 96 = 0

Method. family of lines through an intersection: L₁ + λL₂ = 0

Line: (2x − 3y + 7) + λ(7x + 4y + 2) = 0.
Through (2, 3): (4 − 9 + 7) + λ(14 + 12 + 2) = 0, so 2 + 28λ = 0 and λ = -114.
Multiply by 14: 28x − 42y + 98 − 7x − 4y − 2 = 0, i.e. 21x − 46y + 96 = 0.
Check (2, 3): 42 − 138 + 96 = 0. (holds)

Shortcut. Test the intersection point (−3429, 4529) in the options: all four pass through (2, 3), but only 21x − 46y + 96 = 0 also passes through the intersection; or compare slopes, the required slope being 2146.

Answer: (b) x − y = 0

Method. vertices from the given sides and diagonal

A = AB ∩ AD: y = 0, x = 0, so A = (0, 0).
B = AB ∩ BD: y = 0, x + y = 4, so B = (4, 0); D = AD ∩ BD: D = (0, 4).
C = B + D − A = (4, 4).
AC through (0, 0) and (4, 4): y = x, i.e. x − y = 0.

Answer: (c) 4 square units

Method. area of a square = (diagonal)²/⁠2

B and D are opposite vertices, so BD is a diagonal.
BD = ⁠√((4⁠ −⁠ 2)² + (1 −⁠ 3)²) = ⁠√8.⁠
Area = BD²2 = 82 = 4 square units.

Answer: (d) 4 − 2⁠√3⁠

Method. equal sides OA = OB give p = q, then AB = OA

Let O(0, 0), A(p, 1), B(1, q).
OA = OB: p² + 1 = 1 + q², so p = q (both positive).
AB² = (1 − p)² + (q − 1)² = 2(1 − p)²; set equal to OA² = p² + 1.
2 − 4p + 2p² = p² + 1 ⟹ p² - 4p + 1 = 0 ⟹ p = 2 ± ⁠√3.⁠
0 < p < 1 gives p = 2 - ⁠√3,⁠ so p + q = 4 − 2⁠√3.⁠

Answer: (d) 3x − 4y + 30 = 0

Method. BC is the line through D perpendicular to AD

In an equilateral triangle the altitude AD is perpendicular to BC, and D lies on BC.
Direction of AD: D − A = (-3, 4), which is a normal to BC.
BC: -3x + 4y = -3(-2) + 4(6) = 30, i.e. 3x − 4y + 30 = 0.

No correct option. This question is defective as printed in UPSC's paper.

Method. perpendicular diagonals

Slopes of the diagonals: x − 2y = 1 has slope 12; 4x + 2y = 3 has slope -2.
Product = -1, so the diagonals are perpendicular; nothing else about ABCD is given.
A rhombus has perpendicular diagonals, so ABCD may be a rhombus.
But a square also has perpendicular diagonals, and a square is a rectangle, a parallelogram and a cyclic quadrilateral; every rhombus is a parallelogram too.

The diagonals are perpendicular, which every rhombus satisfies but so does a square, and a square is also a rectangle, a cyclic quadrilateral and a parallelogram; with 'may be', all four options are possible. The conventional reading, a quadrilateral characterised by perpendicular diagonals, gives (d), rhombus.

Answer: (b) 10

Method. diagonals of a parallelogram bisect each other

Taking the vertices in the order P, Q, R, S, the diagonals PR and QS have the same mid-point.
P + R = Q + S ⟹ S = (2 + 10 − 8, 4 + 14 − 12) = (4, 6).
x + y = 10.
(The other orderings give S = (0, 2) or (16, 22), with x + y = 2 or 38, neither an option.)

Answer: (c) (1, 1)

Method. DC ∥ AB fixes the y-coordinate of D; isosceles means AD = BC

AB is horizontal (y = 3), so DC is horizontal through C(5, 1): D = (d, 1).
Isosceles trapezium: AD = BC = ⁠√((5⁠ −⁠ 4)² + (1 −⁠ 3)²) = ⁠√5.⁠
(d − 2)² + 4 = 5 ⟹ d = 3 or d = 1.
d = 3 makes AD ∥ BC (both along (1, -2)), a parallelogram, not a trapezium.
So D = (1, 1).

Answer: (b) (3, 7/⁠3)

Method. intersection of the diagonals AC and BD

Vertices: A(2, 3), B(4, 3), C(5, 1), D(1, 1); the axis of symmetry is x = 3, so the diagonals meet on x = 3.
AC: (x, y) = (2 + 3t, 3 − 2t); x = 3 gives t = 13, y = 3 − 23 = 73.
Check on BD: (4 − 3s, 3 − 2s); x = 3 gives s = 13, y = 73.
Intersection: (3, 73).

Answer: (a) 6.05

Method. distance between parallel lines gives the side of the square

The lines are parallel: 4x + 2y + 5 = 0 is 2x + y + 52 = 0.
Side = distance between 2x + y − 3 = 0 and 2x + y + 52 = 0 = |-3 − 52|⁠√(4⁠ + 1) = 112⁠√5.⁠
Area = 12120 = 6.05 square units.

Answer: (c) 3x − 2y = 0

Method. vertices of the quadrilateral; diagonal through origin to the opposite vertex

The quadrilateral is bounded by x ≥ 0, y ≥ 0, x + y ≤ 1, 6x + y ≤ 3.
Vertices: O(0, 0); 6x + y = 3 meets y = 0 at (12, 0); x + y = 1 meets x = 0 at (0, 1); x + y = 1 and 6x + y = 3 meet where 5x = 2: (25, 35).
The vertex opposite O is (25, 35); diagonal: y = 32x, i.e. 3x − 2y = 0.

Answer: (d) 2x + y − 1 = 0

Method. diagonal joining the two vertices on the axes

From the vertices O(0, 0), (12, 0), (25, 35), (0, 1), the other diagonal joins (12, 0) and (0, 1).
Intercept form: x1/⁠2 + y1 = 1, i.e. 2x + y − 1 = 0.

Answer: (c) Both 1 and 2

Method. third vertex of an equilateral triangle via the perpendicular bisector

Side = ⁠√(2²⁠ + 2²) = 2⁠√2;⁠ mid-point of the base is (1, 1); height = ⁠√32⁠ × 2⁠√2⁠ = ⁠√6.⁠
The third vertex lies on the perpendicular bisector, direction 1⁠√2(1,⁠ -1), at distance ⁠√6⁠ from (1, 1).
Third vertex = (1, 1) ± ⁠√3(1,⁠ -1) = (1 + ⁠√3,⁠ 1 - ⁠√3)⁠ or (1 - ⁠√3,⁠ 1 + ⁠√3):⁠ both coordinates irrational, so statement 1 is correct.
Area = ⁠√34(2√2)²⁠ = 2⁠√3,⁠ irrational, so statement 2 is correct.
Answer: both.

Answer: (d) 2⁠√3⁠

Method. third vertex from the perpendicular bisector

From the shared data, the third vertex is (1 + ⁠√3,⁠ 1 - ⁠√3)⁠ or (1 - ⁠√3,⁠ 1 + ⁠√3).⁠
Difference of coordinates = (1 + ⁠√3)⁠ − (1 - ⁠√3)⁠ = 2⁠√3⁠ (in magnitude, for either vertex).

Answer: (c) 2x − 3y + 8 = 0

Method. fourth vertex D = A + C − B, then two-point form

Diagonals bisect each other, so D = A + C − B = (1 + 3 + 1, 3 + 5 − 2) = (5, 6).
BD through B(-1, 2) and D(5, 6): slope 6 − 25 + 1 = 23.
y − 2 = 23(x + 1) gives 3y − 6 = 2x + 2, i.e. 2x − 3y + 8 = 0.

Answer: (c) 2 square units

Method. area of parallelogram = |cross product of adjacent sides|

Adjacent sides from B: BA = (1 + 1, 3 − 2) = (2, 1) and BC = (3 + 1, 5 − 2) = (4, 3).
Area = |2 × 3 − 1 × 4| = 2 square units.

Shortcut. Area = 2 × area of triangle ABC = |x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)| = |1(-3) + (-1)(2) + 3(1)| = 2.

Answer: (d) 2x − y − 5 = 0

Method. altitude: through the vertex, perpendicular to the opposite side

B = AB ∩ BC: x = 2, y = -1, so B(2, -1).
AC: x + 2y − 4 = 0 has slope -12, so the altitude has slope 2.
Altitude: y + 1 = 2(x − 2), i.e. 2x − y − 5 = 0.

Answer: (a) (4, 0)

Method. right-angled triangle: circumcentre is the mid-point of the hypotenuse

AB: x = 2 and BC: y = -1 are perpendicular, so the right angle is at B(2, -1).
A = AB ∩ CA: x = 2, 2 + 2y = 4, so A(2, 1). C = BC ∩ CA: y = -1, x − 2 = 4, so C(6, -1).
Circumcentre = mid-point of hypotenuse AC = (2 + 62, 1 − 12) = (4, 0).

Answer: (a) (2, 10)

Method. diagonals of a parallelogram bisect each other: D = A + C − B

Let A(-3, 4), B(0, -4), C(5, 2) be consecutive vertices.
Mid-point of AC = mid-point of BD, so D = A + C − B.
D = (-3 + 5 − 0, 4 + 2 + 4) = (2, 10).

Answer: (a) 2x + y = 0

Method. diagonals of a square are perpendicular and pass through the centre

Diagonal AC passes through O(0, 0) and A(4, 2): slope 24 = 12.
BD is perpendicular to AC and also passes through O: slope -2.
BD: y = -2x, i.e. 2x + y = 0.

Answer: (c) 2x − 3y = 5

Method. diagonals of a square are perpendicular; the given vertex is not on the given diagonal

Check (1, -1) on 3x + 2y = 5: 3 − 2 = 1 ≠ 5, so this vertex lies on the other diagonal.
The diagonals of a square are perpendicular, so the other diagonal has the form 2x − 3y = c.
Through (1, -1): c = 2(1) − 3(-1) = 5.
Other diagonal: 2x − 3y = 5.

Answer: (c) 1

Method. right-angled triangle: circumcentre is the mid-point of the hypotenuse

x + 2 = 0 and y + 2 = 0 are perpendicular and meet at A(-2, -2), so the right angle is at A.
kx + y + 2 = 0 meets x = -2 at B(-2, 2k − 2) and y = -2 at C(0, -2) (k ≠ 0).
Circumcentre = mid-point of BC = (-2 + 02, 2k − 2 − 22) = (-1, k − 2).
k − 2 = -1 gives k = 1.

Answer: (c) 16 square units

Method. side of the square = distance between the parallel sides

Distance between 3x − 4y − 5 = 0 and 3x − 4y + 15 = 0 is |-5 − 15|⁠√(3²⁠ + 4²) = 205 = 4.
This is the side of the square, so area = 4² = 16 square units.

Answer: (c) (1, 1)

Method. the third vertex is at distance AB from both given vertices

A(-1, -1), B(-⁠√3,⁠ ⁠√3):⁠ AB² = (1 - ⁠√3)²⁠ + (1 + ⁠√3)²⁠ = 8.
Test (1, 1): to A, 2² + 2² = 8; to B, (1 + ⁠√3)²⁠ + (1 - ⁠√3)²⁠ = 8. Both equal AB², so it is the third vertex.
The others fail: (-⁠√2,⁠ ⁠√2)⁠ and (⁠√2,⁠ -⁠√2)⁠ are at distance² 6 from A, and (1, -1) at distance² 4.
The third vertex may lie at (1, 1).

Answer: (d) -4

Method. the other diagonal of a rectangle passes through the midpoint of the given diagonal

The other two vertices are the ends of the other diagonal, so y = 2x + c is that diagonal.
The diagonals bisect each other at the midpoint of (1, 3) and (5, 1): (3, 2).
2 = 2.3 + c, so c = -4.

Answer: (a) isosceles

Method. find the vertices and compare the side lengths

x + y = 4 and 3x + y = 4 meet at (0, 4); x + y = 4 and x + 3y = 4 meet at (4, 0); 3x + y = 4 and x + 3y = 4 meet at (1, 1).
Sides: (0, 4) to (4, 0) = 4⁠√2;⁠ (0, 4) to (1, 1) = ⁠√10;⁠ (4, 0) to (1, 1) = ⁠√10.⁠
Two sides are equal; 10 + 10 ≠ 32, so there is no right angle.
The triangle is isosceles.

Answer: (a) (1, 2)

Method. diagonals of a parallelogram bisect each other

Let A(-2, -1), B(1, 0), C(4, 3) be consecutive vertices and D the fourth.
Midpoint of AC = midpoint of BD, so D = A + C − B.
D = (-2 + 4 − 1, -1 + 3 − 0) = (1, 2).