Question

What will be the value of K if 3 points (7, -2), (5, 1) and (3, K) are co-linear ?

Answer 1

✬ The value of k is 4 ✬

Step-by-step explanation:

Given:-

• Collinear points are (7,-2),(5,1) and (3,K).

To find:-

• Value of k

Required Formula:-

• y₂-y₁/x₂-x₁ = y₃-y₂/x₃-x₂

Solution:-

• Let the points be A(x1,y1) = (7,-2) ,B (x2,y2) = (5,1) and C(x3,y3) = (3,k)

Then, applying the values,

$\\:\implies\sf{\frac{y_2-y_1}{x_2-x_1}=\frac{y_3-y_2}{x_3-x_2}} \\ \\ :\implies\sf{ \frac{1 - ( - 2)}{5 - 7} = \frac{k - 1}{3 - 5} } \\ \\ :\implies\sf{ \frac{3}{ - 2} = \frac{k - 1}{ - 2} }$

Cross multiplying the values,

$\\ :\implies\sf{3( - 2) = - 2(k - 1)} \\ \\ :\implies\sf{ - 6 = - 2k + 2} \\ \\ :\implies\sf{ -6 - 2 = - 2k} \\ \\ :\implies\sf{ - 8 = - 2k} \\ \\ :\implies\sf{ - k = \frac{ - 8}{2} } \\ \\ :\implies\sf{ \cancel{ - } \: k = \cancel{ -} \: 4} \\ \\ :\implies \underline{\boxed{\frak{k = 4}}} \: \pink{ \bigstar}$

• Hence,The value of k is 4.

_________________________________

Verification:-

• Now, Putting the value of k in the place of k.

$\\ :\implies\sf{ \frac{1 - ( - 2)}{5 - 7} = \frac{4 - 1}{3 - 5} } \\ \\ :\implies\sf{ \frac{3}{ - 2} = \frac{3}{ - 2} }$

• Hence,Verified.

_________________________________