SOLVING FOR UNKNOWN WITH COLLINEAR POINTS

Problem 1 :

Find the value of 𝑝 for which the points

(−1, 3), (2, 𝑝) and (5, −1)

are collinear.

Solution :

Let the points be A (−1, 3) B (2, 𝑝) and C (5, −1)

x₁ = -1, y₁ = 3, x₂ = 2, y₂ = p, x₃ = 5, y₃ = -1

-1[p - (-1)] + 2[-1 - 3] + 5[3 - p] = 0

-1(p+1)+2(-4)+5[3-p] = 0

-p-1-8+15-5p = 0

-6p+15-9 = 0

-6p+6 = 0

-6p = -6

p = -1

So, the value of p is -1.

Problem 2 :

Find the value of 𝑘 if 𝑃(4, −2) is the mid point of the line segment joining the points 𝐴(5𝑘, 3) and 𝐵(−𝑘, −7).

Solution :

P is the midpoint of line joining the point A and B.

x₁ = 5k, y₁ = 3, x₂ = -k, y₂ = -7

Equating x coordinate,

(5k - k)/2 = 4

4k/2 = 4

2k = 4

k = 4/2

k = 2

So, the value of k is 2.

Problem 3 :

Find the value of 𝑎 so that the point (3, 𝑎) lies on the line represented by 2𝑥 − 3𝑦 = 5

Solution :

(3, a) lies on the line 2x - 3y = 5

When x = 3 and y = a

2(3) - 3a = 5

6 - 3a = 5

-3a = 5 - 6

-3a = -1

a = 1/3

Problem 4 :

If the mid-point of the line segment joining the points

𝑃(6, 𝑏 − 2) and 𝑄(−2, 4) is (2, −3)

find the value of b.

Solution :

Midpoint of the line segment joining the points P and Q is (2, -3)

Equating y-coordinates :

(b - 2 + 4)/2 = - 3

(b + 2)/2 = -3

b + 2 = -6

b = -6 - 2

b = -8

So, the value of b is -8.

Problem 5 :

For what value of 𝑘 are the points (8, 1), (3, −2𝑘) and (𝑘, −5) collinear?

Solution :

Let the given point be A (8, 1), B (3, −2𝑘) and C (𝑘, −5)

x₁ = 8, y₁ = 1, x₂ = 3, y₂ = -2k, x₃ = k, y₃ = -5

8(-2k - (-5)) + 3(-5 - 1) + k(1 - (-2k)) = 0

8(-2k + 5) + 3(-6) + k(1 + 2k) = 0

-16k + 40 - 18 + k + 2k² = 0

2k² - 15k + 22 = 0

2k² - 11k - 4k + 22 = 0

k(2k - 11) - 2(2k - 11) = 0

(k - 2) (2k - 11) = 0

k = 2 and k = 11/2

So, the value of k is 2 and 11/2.

Problem 6 :

If the points 𝐴(4, 3) and 𝐵(𝑥, 5) are on the circle with the center 𝑂(2, 3), find the value of 𝑥.

Solution :

Center of the circle = (2, 3)

Distance from the center of the circle to any point on the boundary is the same and we know it is radius.

Distance between two points :

= √(x₂ - x₁)² + (y₂-y₁)²

√(2 - 4)² + (3 - 3)² = √(2 - x)² + (3 - 5)²

Take square on both sides.

4 + 0 = (2 - x)² + 4

0 = (2 - x)²

2 - x = 0

x = 2

So, the value of x is 2.