HOW TO FIND MISSING COORDINATE WHEN GIVEN POINTS ARE COLLINEAR

Problem 1 :

For what value of x are the points

A(-3, 12), B(7, 6) and C(x, 9)

collinear ?

a)1       b) -1     c) 2      d) -2

Solution :

Given points A(-3, 12), B(7, 6) and C(x, 9) are collinear.

Slope = (y₂ – y₁)/(x₂ – x₁)

(1)  = (2)

-3/5 = 3/(x – 7)

-3(x – 7) = 5(-3)

-3x + 21 = -15

-3x = -15 – 21

-3x = -6

x = 2

So, the value of x is 2. Hence, option c) is correct.

Problem 2 :

For what value of y are the points

A(1, 4), B(3, y) and C(-3, 16)

collinear ?

a)1      b) -1     c) 2     d) -2

Solution :

Given points A(1, 4), B(3, y) and C(-3, 16) are collinear.

Slope = (y₂ – y₁)/(x₂ – x₁)

(1) = (2)

(y – 4)/2 = (16 – y)/-6

-6(y – 4) = 2(16 - y)

-6y + 24 = 32 – 2y

-6y + 2y = 32 – 24

-4y = 8

Dividing 4 on each sides.

y = -2

So, the value of y is -2. Hence, option d) is correct.

Problem 3 :

Find the value of p for which the points

A(-1, 3), B(2, p) and C(5, -1)

collinear ?

a)1      b) -1      c) 2       d) -2

Solution :

Given points A(-1, 3), B(2, p) and C(5, -1) are collinear.

Slope = (y₂ – y₁)/(x₂ – x₁)

(1)  = (2)

(p - 3)/3 = (-1 – p)/3

3(p – 3) = 3(-1 - p)

3p - 9 = -3 – 3p

3p + 3p = -3 + 9

6p = 6

Dividing p on each sides.

p = 1

So, the value of p is 1. Hence, option a) is correct.

Problem 4 :

If the points

(1, x), (5, 2) and (9, 5)

are collinear then the value of x is

a) 5/2     b) -5/2     c) -1     d) 1

Solution :

Given points A(1, x), B(5, 2) and C(9, 5) are collinear.

(1)  = (2)

(2 – x)/4 = 3/4

4(2 – x) = 3(4)

8 – 4x = 12

-4x = 12 – 8

-4x = 4

x = -1

So, the value of x is -1. Hence, option c) is correct.

Problem 5 :

If the points (-1, 1), (2, p) and (8, 11) are collinear, find the value of p using section formula.

Solution :

Given points, (-1, 1), (2, p) and (8, 11) are collinear.

Let k : 1 be the ratio that divides the line segment joining the points (-1, 1) and (8, 11) at (2, p).

(lx₂ + mx₁)/(l + m), (ly₂ + my₁)/(l + m) = (2, p)

(x₁, y₁) = (-1, 1)

(x₂, y₂) = (8, 11)

(k(8) + 1(-1))/(k + 1), (k(11) + 1(-1))/(k + 1) = (2, p)

(8k - 1)/(k + 1) = 2 --- (1)

(11k - 1)/(k + 1) = p --- (2)

(8k - 1)/(k + 1) = 2

8k - 1 = 2(k + 1)

8k – 1 = 2k + 2

8k – 2k = 2 + 1

6k = 3

k = 3/6

k = 1/2

k = 1/2 substitute in equation (2). 

(11(1/2) – 1)/(1/2 + 1) = p

(11/2 – 1)/(3/2) = p

(9/2)/(3/2) = p

9/3 = p

p = 3

So, the value of p is 3.

Problem 6 :

If the points (2, 3), (4, t) and (6, -3) are collinear, find the value of k using section formula.

Solution :

Given points, (2, 3), (4, t) and (6, -3) are collinear.

Let k : 1 be the ratio that divides the line segment joining the points (2, 3) and (4, k) at (6, -3).

(lx₂ + mx₁)/(l + m), (ly₂ + my₁)/(l + m) = (4, t)

(x₁, y₁) = (2, 3)

(x₂, y₂) = (6, -3)

(k(6) + 1(2))/(k + 1), (k(-3) + 1(3))/(k + 1) = (4, t)

(6k + 2)/(k + 1) = 4 --- (1)

(-3k + 3)/(k + 1) = t --- (2)

(6k + 2)/(k + 1) = 4

6k + 2 = 4(k + 1)

6k + 2 = 4k + 4

6k – 4k = 4 - 2

2k = 2

k = 1

So, the value of k is 1.

Applying the value of k in (2), we get

(-3 + 3)/(3 + 1) = t

t = 0