Analytical Geometry — Lines, Circles, Conics & Three Dimensions — Practice

Question 1

The distance between the points A(3, -2) and B(-1, 1) is:
(a) 5  (b) 7  (c) √7  (d) 25

Accepted Answer

PQ = √[(-1-3)² + (1-(-2))²] = √[16+9] = √25 = 5.

Answer: (a) 5. Trap (b) 7 comes from adding the absolute coordinate differences directly (|-4|+|3|=7) instead of using the distance formula's squares. Trap (c) √7 comes from forgetting to square the differences before summing: √(4+3)=√7. Trap (d) 25 is the squared distance (d²) with the final square root forgotten.

Question 2

The point dividing the line segment joining A(2, 3) and B(7, 8) internally in the ratio 2:3 is:
(a) (4, 5)  (b) (5, 6)  (c) (-8, -7)  (d) (9/2, 11/2)

Accepted Answer

x = (2×7+3×2)/(2+3) = (14+6)/5 = 4. y = (2×8+3×3)/5 = (16+9)/5 = 5.

Answer: (a) (4, 5). Trap (b) (5, 6) comes from swapping the ratio order — using 3:2 instead of 2:3: (3×7+2×2)/5=5, (3×8+2×3)/5=6. Trap (c) (-8, -7) comes from applying the EXTERNAL section formula instead of internal: (2×7-3×2)/(2-3)=-8, (2×8-3×3)/(2-3)=-7. Trap (d) (9/2, 11/2) is simply the midpoint of AB, ignoring the given ratio entirely.

Question 3

The area of the triangle with vertices (1, 2), (3, 4) and (5, 10) is:
(a) 4  (b) 8  (c) 9  (d) 2

Accepted Answer

Δ = ½|1(4-10) + 3(10-2) + 5(2-4)| = ½|1(-6)+3(8)+5(-2)| = ½|-6+24-10| = ½(8) = 4.

Answer: (a) 4. Trap (b) 8 forgets the ½ factor, reporting the shoelace sum directly. Trap (c) 9 drops the third vertex's contribution entirely: ½|1(-6)+3(8)| = ½(18) = 9. Trap (d) 2 applies an extra, incorrect ½ (quartering instead of halving): 8/4=2.

Question 4

The equation of the line passing through (2, -1) and (5, 8) is:
(a) 3x - y - 7 = 0  (b) 3x - y - 5 = 0  (c) x - 3y - 5 = 0  (d) 3x + y - 5 = 0

Accepted Answer

Slope m = (8-(-1))/(5-2) = 9/3 = 3. Point-slope: y+1=3(x-2) ⇒ y+1=3x-6 ⇒ y=3x-7 ⇒ 3x-y-7=0.

Answer: (a) 3x - y - 7 = 0. Trap (b) 3x-y-5=0 comes from a transposition sign slip: writing y=3x-6+1=3x-5 instead of correctly y=3x-6-1=3x-7. Trap (c) x-3y-5=0 comes from using the reciprocal slope 1/3 instead of 3 (swapping which coordinate difference goes on top). Trap (d) 3x+y-5=0 comes from flipping the slope's sign to -3 (as if reading the points in reverse order incorrectly), giving y+1=-3(x-2).

Question 5

The acute angle between the lines √3x - y + 1 = 0 and x - √3y + 1 = 0 is:
(a) 30°  (b) 60°  (c) 90°  (d) 150°

Accepted Answer

Slopes: m1=√3, m2=1/√3 (from slope=−a/b for each line). tanθ = |(√3−1/√3)/(1+√3·1/√3)| = |(2/√3)/2| = 1/√3 ⇒ θ=30°.

Answer: (a) 30°. Trap (b) 60° confuses tan30°=1/√3 with tan60°=√3 — a standard-angle table mix-up, reading the correct ratio but recalling the wrong angle. Trap (c) 90° comes from computing the denominator as 1−m1m2 instead of 1+m1m2 — here m1m2=1, so this sign slip zeroes the denominator, misread as the lines being perpendicular. Trap (d) 150° is the obtuse supplementary angle, obtained by dropping the absolute value in the tanθ formula — both 30° and 150° are genuine angles between the lines, but only the acute one is asked for.

Question 6

The distance of the point (1, 5) from the line 3x - 4y + 10 = 0 is:
(a) 7/5  (b) -7/5  (c) 7  (d) 7/25

Accepted Answer

d = |3(1)-4(5)+10|/√(9+16) = |3-20+10|/5 = |-7|/5 = 7/5.

Answer: (a) 7/5. Trap (b) -7/5 forgets the absolute value, reporting the signed numerator directly. Trap (c) 7 forgets to divide by √(a²+b²) entirely. Trap (d) 7/25 divides by (a²+b²)=25 instead of √(a²+b²)=5 — squaring the denominator instead of taking its square root.

Question 7

The three lines x + y = 1, 2x + 3y = 5 and x - ky = 3 are concurrent for k equal to:
(a) -5/3  (b) 5/3  (c) -3/5  (d) 3/5

Accepted Answer

Intersection of the first two: from x=1-y, 2(1-y)+3y=5 ⇒ y=3, x=-2. Substituting (-2,3) into the third line: -2-3k=3 ⇒ -3k=5 ⇒ k=-5/3. (Cross-check via the determinant |1 1 -1; 2 3 -5; 1 -k -3| = -5-3k = 0 gives the same k=-5/3.)

Answer: (a) -5/3. Trap (b) 5/3 comes from dropping the negative sign on the ky term — treating the third line as x+ky=3 instead of x-ky=3: -2+3k=3 ⇒ k=5/3. Traps (c) -3/5 and (d) 3/5 are the reciprocal of the correct answer (with and without the sign flip), from inverting the fraction while isolating k from -3k=5.

Question 8

If the lines represented by 3x² + 7xy + 2y² = 0 are inclined to each other at angle θ, then tanθ equals:
(a) 1  (b) 5  (c) -1  (d) 1/2

Accepted Answer

Here a=3, b=2, 2h=7 so h=3.5. tanθ = 2√(h²-ab)/(a+b) = 2√(12.25-6)/5 = 2√6.25/5 = 2(2.5)/5 = 5/5 = 1. (Cross-check by factoring: 3x²+7xy+2y²=(3x+y)(x+2y), slopes -3 and -1/2; tanθ=|(-3+1/2)/(1+1.5)|=|-2.5/2.5|=1.)

Answer: (a) 1. Trap (b) 5 forgets to divide by (a+b), reporting the numerator 2√(h²-ab) alone. Trap (c) -1 is the tangent of the OTHER (obtuse, 135°) angle between the same pair of lines — since tan(180°-θ)=-tanθ, this is a genuine but wrongly-selected value. Trap (d) 1/2 forgets the factor of 2 in the numerator, computing √(h²-ab)/(a+b) = 2.5/5 = 1/2 instead.

Question 9

The centre and radius of the circle x² + y² - 6x + 8y - 11 = 0 are:
(a) (3, -4), 6  (b) (-3, 4), 6  (c) (3, -4), √14  (d) (3, 4), 6

Accepted Answer

2g=-6 ⇒ g=-3; 2f=8 ⇒ f=4; c=-11. Centre = (-g,-f) = (3,-4). Radius = √(g²+f²-c) = √(9+16-(-11)) = √36 = 6.

Answer: (a) (3, -4), 6. Trap (b) (-3, 4), 6 forgets to negate g and f, using (g,f) directly instead of (-g,-f). Trap (c) (3, -4), √14 uses c=+11 instead of the actual c=-11 when computing the radius: √(9+16-11)=√14. Trap (d) (3, 4), 6 negates only g (getting the x-coordinate right) but forgets to negate f.

Question 10

For what value(s) of c is the line y = x + c tangent to the circle x² + y² = 8?
(a) c = ±4  (b) c = ±2√2  (c) c = ±8  (d) c = ±2

Accepted Answer

Tangency: distance from centre (0,0) to x-y+c=0 equals radius √8=2√2. |c|/√(1+1) = 2√2 ⇒ |c| = 2√2×√2 = 4.

Answer: (a) c = ±4. Trap (b) c=±2√2 skips the √(a²+b²)=√2 division entirely, setting |c| equal to the radius directly. Trap (c) c=±8 comes from mistakenly doubling the radius before applying the tangent condition (treating distance = 2r instead of distance = r). Trap (d) c=±2 comes from mis-simplifying √8 as √2 instead of the correct 2√2, giving |c|/√2=√2 ⇒ |c|=2.

Question 11

For the parabola y² = 12x, the focus and the length of the latus rectum are:
(a) (3, 0), 12  (b) (3, 0), 6  (c) (12, 0), 12  (d) (0, 3), 12

Accepted Answer

4a=12 ⇒ a=3. Focus = (a,0) = (3,0). Latus rectum = 4a = 12.

Answer: (a) (3, 0), 12. Trap (b) (3, 0), 6 computes latus rectum as 2a instead of 4a — a confusion with the ellipse/hyperbola latus rectum shape (2b²/a). Trap (c) (12, 0), 12 mistakes the coefficient 12 itself for the focus's x-coordinate, skipping the step of solving 4a=12 for a first. Trap (d) (0, 3), 12 swaps the focus's coordinates, confusing this rightward-opening y²=4ax form with the upward-opening x²=4ay form.

Question 12

The eccentricity of the ellipse 9x² + 25y² = 225 is:
(a) 4/5  (b) 3/5  (c) 4/3  (d) 16/25

Accepted Answer

Divide by 225: x²/25 + y²/9 = 1, so a²=25, b²=9 (a>b, major axis along x). c²=a²-b²=16 ⇒ c=4. e=c/a=4/5.

Answer: (a) 4/5. Trap (b) 3/5 computes b/a instead of c/a — a common confusion between the eccentricity ratio and the axis ratio. Trap (c) 4/3 comes from wrongly taking a²=9 (the smaller denominator) instead of a²=25, giving e=c/√9=4/3 — an impossible value (>1) that should be a red flag for an ellipse. Trap (d) 16/25 forgets to take the square root, reporting c²/a² instead of c/a.

Question 13

The distance between the points P(2, 3, 4) and Q(5, 7, 16) is:
(a) 13  (b) 5  (c) 12  (d) 169

Accepted Answer

PQ = √[(5-2)²+(7-3)²+(16-4)²] = √[9+16+144] = √169 = 13.

Answer: (a) 13. Trap (b) 5 drops the z-term entirely and uses only the 2D distance formula: √(9+16)=√25=5 — a classic habit-driven error when moving from 2D to 3D. Trap (c) 12 mistakes the largest single coordinate difference (Δz=12) for the full 3D distance. Trap (d) 169 is the squared distance PQ² with the final square root forgotten.

Question 14

If a line has direction ratios (1, -2, 2), its direction cosines are:
(a) (1/3, -2/3, 2/3)  (b) (1, -2, 2)  (c) (1/9, -2/9, 2/9)  (d) (1/√3, -2/√3, 2/√3)

Accepted Answer

Magnitude = √(1+4+4) = √9 = 3. Direction cosines = (1/3, -2/3, 2/3).

Answer: (a) (1/3, -2/3, 2/3). Trap (b) (1, -2, 2) reports the direction ratios directly as if they were already direction cosines, skipping the normalisation step entirely. Trap (c) (1/9, -2/9, 2/9) divides by the magnitude SQUARED (9) instead of the magnitude itself (3). Trap (d) (1/√3, -2/√3, 2/√3) comes from an arithmetic slip while summing the squares — e.g. treating the middle term as -2 instead of (-2)²=4, giving 1-2+4=3 and a magnitude of √3 instead of the correct √9=3.

Question 15

The distance of the point (2, 3, -1) from the plane x + 2y - 2z - 4 = 0 is:
(a) 2  (b) 6  (c) 2/3  (d) 18

Accepted Answer

|1(2)+2(3)+(-2)(-1)+(-4)| = |2+6+2-4| = 6. Denominator = √(1+4+4) = 3. Distance = 6/3 = 2.

Answer: (a) 2. Trap (b) 6 forgets to divide by √(a²+b²+c²), reporting the numerator alone. Trap (c) 2/3 divides by (a²+b²+c²)=9 instead of √(a²+b²+c²)=3, squaring the denominator instead of taking its square root: 6/9=2/3. Trap (d) 18 inverts the operation, multiplying by 3 instead of dividing: 6×3=18.

Question 16

The angle between the planes x + y + 2z = 9 and 2x - y + z = 15 is:
(a) 60°  (b) 30°  (c) 90°  (d) 120°

Accepted Answer

Normals n1=(1,1,2), n2=(2,-1,1). n1·n2 = 2-1+2=3. |n1|=|n2|=√6. cosθ = 3/(√6×√6) = 3/6 = 1/2 ⇒ θ=60°.

Answer: (a) 60°. Trap (b) 30° confuses cosθ=1/2 (giving 60°) with sinθ=1/2 (giving 30°) — a cos⁻¹ vs sin⁻¹ mix-up on the identical ratio. Trap (c) 90° comes from the false leap that since the normals aren't scalar multiples of each other (not parallel), the planes must be perpendicular — only a zero dot product actually proves perpendicularity. Trap (d) 120° comes from a sign error that flips the entire dot product to -3, giving cosθ=-1/2 — the supplementary (obtuse) angle instead of the correct one.

Practice — Analytical Geometry — Lines, Circles, Conics & Three Dimensions

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