Solutions — Parametric Equations of Lines

Q1 — Write Parametric Equations

The line passes through A = (3, −1) with direction d = 2i + 5j.

Vector form: r = (3i − j) + t(2i + 5j)

Parametric equations: x = 3 + 2t, y = −1 + 5t

Cartesian form: Eliminating t: t = (x − 3)/2 = (y + 1)/5, giving 5(x − 3) = 2(y + 1), i.e. 5x − 2y = 17.

Q2 — Line Through Two Points

P = (−1, 4), Q = (3, 2).

Direction vector: PQ = Q − P = (3−(−1))i + (2−4)j = 4i − 2j

Simplify by dividing by 2: d = 2i − j

Parametric equations (using P as base point):

x = −1 + 2t, y = 4 − t

Check at t=0: P=(−1,4) ✓. At t=2: Q=(3,2) ✓ (using d=(2,−1), so t=2 from −1+2×2=3).

Equivalent form using Q: x=3+2t, y=2−t. Both are correct.

Q3 — Points on a Line

Line: r = (2i + 3j) + t(i + 2j). Parametric: x = 2 + t, y = 3 + 2t.

(a) t = −1: x = 2 + (−1) = 1, y = 3 + 2(−1) = 1. Point: (1, 1)

(b) t = 3: x = 2 + 3 = 5, y = 3 + 6 = 9. Point: (5, 9)

Note: as t increases by 1, the point moves one direction-vector step along the line.

Q4 — Convert to Cartesian Form

Parametric: x = 1 + 3t, y = 4 − t.

Step 1: Solve the first equation for t: t = (x − 1)/3

Step 2: Substitute into the second equation:

y = 4 − (x − 1)/3 ⇒ 3y = 12 − (x − 1) = 13 − x

Cartesian form: x + 3y = 13

Verify with a point: at t=0, point is (1,4). Check: 1+12=13 ✓. At t=1, point=(4,3). Check: 4+9=13 ✓.

Q5 — Parallel, Coincident or Intersecting?

ℓ₁: direction (4, 6) = 2(2, 3). ℓ₂: direction (2, 3).

Step 1 — Check directions: (4,6) = 2×(2,3), so the lines are parallel or coincident.

Step 2 — Check coincidence: Does (1,2) from ℓ₁ lie on ℓ₂?

(1, 2) = (3, 5) + s(2, 3)

x: 1 = 3 + 2s ⇒ s = −1

Check y: 5 + 3(−1) = 2 ✓

Since the point (1,2) from ℓ₁ lies on ℓ₂ at s = −1, the lines are coincident (the same line).

Q6 — Find the Intersection

ℓ₁: x=1+t, y=2t. ℓ₂: x=3−s, y=2+s.

Set equal component-wise:

x: 1 + t = 3 − s ⇒ t + s = 2 ...(1)

y: 2t = 2 + s ⇒ 2t − s = 2 ...(2)

Add (1) and (2): 3t = 4 ⇒ t = 4/3. Then s = 2 − 4/3 = 2/3.

Intersection: x = 1 + 4/3 = 7/3, y = 2(4/3) = 8/3. Point: (7/3, 8/3)

Verify on ℓ₂: (3 − 2/3, 2 + 2/3) = (7/3, 8/3) ✓

Q7 — Closest Point to the Origin

Line ℓ: x = 2 + 3t, y = 1 + 4t. Direction: d = (3, 4).

A general point on ℓ: Q = (2 + 3t, 1 + 4t).

Vector from origin to Q: OQ = (2 + 3t)i + (1 + 4t)j

For closest point, OQ ⊥ d, so OQ · d = 0:

3(2 + 3t) + 4(1 + 4t) = 0

6 + 9t + 4 + 16t = 0 ⇒ 25t = −10 ⇒ t = −2/5

Closest point: x = 2 + 3(−2/5) = 2 − 6/5 = 4/5; y = 1 + 4(−2/5) = 1 − 8/5 = −3/5

Closest point: (4/5, −3/5). Distance from origin: √(16/25 + 9/25) = √(25/25) = 1 unit.

Q8 — Does a Point Lie on the Line?

A = (2, 5), B = (6, 3). Direction: AB = (4, −2). Simplify: d = (2, −1).

Parametric equations (base point A):

x = 2 + 2t, y = 5 − t

Test C = (14, −1):

x: 14 = 2 + 2t ⇒ t = 6

Check y: 5 − 6 = −1 ✓

Yes, C lies on the line at t = 6. Since t = 6 > 1 (where t=0 is A and t=2 is B using d=(2,−1)), C lies beyond B in the direction from A.

Q9 — Intersection and Triangle Area

ℓ₁: x=t, y=t. ℓ₂: x=4−s, y=2s.

Finding P:

x: t = 4 − s ...(1)

y: t = 2s ...(2)

Substituting (2) into (1): 2s = 4 − s ⇒ 3s = 4 ⇒ s = 4/3, t = 8/3.

P = (8/3, 8/3)

Triangle OAP area:

O = (0,0), A = (4,0), P = (8/3, 8/3).

Base OA lies along the x-axis with length 4. Height = y-coordinate of P = 8/3.

Area = ½ × base × height = ½ × 4 × 8/3 = 16/3 square units

Q10 — Foot of the Perpendicular

Line ℓ: r = (1, 2) + t(2, 1). General point: Q = (1 + 2t, 2 + t). Direction: d = (2, 1).

PQ = Q − P = (1 + 2t − 5)i + (2 + t − 5)j = (2t − 4)i + (t − 3)j

For perpendicularity: PQ · d = 0:

(2t − 4)(2) + (t − 3)(1) = 0

4t − 8 + t − 3 = 0 ⇒ 5t = 11 ⇒ t = 11/5

Foot of perpendicular: Q = (1 + 22/5, 2 + 11/5) = (27/5, 21/5)

Verify: PQ = (27/5 − 5)i + (21/5 − 5)j = (2/5)i − (4/5)j. PQ · d = (2/5)(2) + (−4/5)(1) = 4/5 − 4/5 = 0 ✓