Question

In each of the following find the value of 'k', for which the points are collinear. (7,-2), (5, 1), (3, k)​

Answer 1

Answer

The value of k is 4

Step by step explanation

Let us consider A(7,-2) , B (5 ,1) and C(3 , k)

Given the points A , B and C are collinear so

the gradients of the line AB will be equals to the gradient of the line BC

Gradient of AB ,

m₁ =(y₂ - y₁)/(x₂ - x₁)

⇒m₁ = {1 - (-2)}/(5 -7)

⇒m₁= -3/2

ᵃnd the gradient of BC,

m₂ = (y₂ - y₁)/(x₂ - x₁)

⇒m₂ =(k - 1)/(3 - 5)

⇒ m₂ = -(k -1)/2

since the gradients of AB and BC are equal so

m₁ = m₂

⇒-3/2 = -(k -1)/2

⇒-3 = -(k - 1)

⇒3 = k - 1

⇒k = 4

Answer 2

$\huge{\boxed{\orange{\bold{Answer}}}}$

The value of "k" = 4

$\huge{\boxed{\bold{\orange{Solution}}}}$

Let us assume that A(7,-2); B(5,1) and C(3,k).

Here, A, B and C are collinear show the gradient of the line AB equals to the gradient of the line BC.

Gradient of line AB

By using this formula, we solve:

$\red{\boxed{ {m}_{1} \: = \: \frac{{y}_{2} - {y}_{1}}{ {x}_{2} - {x}_{1} }}}$

$\blue{= > \: {m}_{1} \: = \frac{1 - ( - 2)}{5 - 7}}$

$\blue{= > \: {m}_{1} = \frac{ - 3}{2}}$

Gradient of line BC:

By using the same formula, we solve:

$\red{\boxed{ {m}_{2} \: = \: \frac{{y}_{2} - {y}_{1}}{ {x}_{2} - {x}_{1} }}}$

$\blue{= > \: {m}_{2} = \frac{k - 1}{3 - 5}}$

$\blue{= > \: {m}_{2} \: = \frac{ (k - 1)}{ - 2}}$

$\blue{= > {m}_{2} = \frac{ - (k - 1)}{2}}$

So therefore, the gradient of line AB and line BC are equal

$\red{\boxed{{m}_{1} = {m}_{2}}}$

$\blue{= > \frac{ - 3}{2} = \frac{ - (k - 1)}{2}}$

$\blue{= > ( - 3) \: = - (k - 1)}$

So, here negative sign cancelled both the sides.

$\blue{= > 3 = k - 1}$

$\blue{= > k = 3 + 1}$

$\blue{\boxed{\bold{\therefore k = 4}}}$

Hence, the value of "k" is 4.