Question

Find the value of k , if the points a[2,3] b[4,k] &-c[6,3] are collinea

Answer 1

Step-by-step explanation:

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Answer 2

$\sf\blue{Correct \ Question}$

$\sf{Find \ the \ value \ of \ k, \ if \ point}$

$\sf{A(2,3); \ B(4,k) \ and \ C(6,3) \ are \ collinear.}$

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$\sf\red{\underline{\underline{Answer:}}}$

$\sf{The \ value \ of \ k \ is \ 3.}$

$\sf\orange{Given:}$

$\sf{Co-ordinates \ of \ collinear \ points \ are:-}$

$\sf{\implies{A(2,3);}}$

$\sf{\implies{B(4,k);}}$

$\sf{\implies{C(6,3).}}$

$\sf\pink{To \ find:}$

$\sf{The \ value \ of \ k.}$

$\sf\green{\underline{\underline{Solution:}}}$

$\sf{If \ points \ are \ collinear \ the \ slope}$

$\sf{of \ lines \ formed \ by \ joining \ points}$

$\sf{will \ be \ equal.}$

$\boxed{\sf{Slope \ of \ line(m)=\frac{y2-y1}{x2-x1}}}$

$\sf{Slope \ of \ line \ AC=\frac{3-3}{6-2}}$

$\sf{\therefore{Slope \ of \ AC=0}}$

$\sf{Slope \ of \ line \ BC=\frac{3-k}{6-4}}$

$\sf{But, \ slope \ of \ line \ AC = Slope \ of \ line \ BC}$

$\sf{\therefore{\frac{3-k}{6-4}=0}}$

$\sf{\therefore{3-k=0}}$

$\sf{\therefore{-k=-3}}$

$\sf{\therefore{k=3}}$

$\sf\purple{\tt{\therefore{The \ value \ of \ k \ is \ 3.}}}$