Respuesta :
Explanation:
[tex]\underline{\underline{\large\bf{Given:-}}}[/tex]
[tex]\red{➤}\:[/tex][tex]\textsf{} [/tex][tex]\sf Points\:(2, 4), (0,3)\: and\; (k, -4) \:are\; collinear[/tex]
[tex]\underline{\underline{\large\bf{To Find:-}}}[/tex]
[tex]\orange{☛}\:[/tex][tex]\textsf{Value of k} [/tex][tex]\sf [/tex]
[tex]\\[/tex]
[tex]\underline{\underline{\large\bf{Solution:-}}}\\[/tex]
Let us consider these points on a line AC such that point B lies in between points A and C as these lines are collinear -
[tex]\\A(2,4) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: B(0,3) \: \: \: \: \: \: \: \: \: \: \: \: C(k,4)\\ \bull \frac{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }{} \bull \frac{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }{} \bull \\[/tex]
If lines are collinear then,
[tex]\green{ \underline { \boxed{ \sf{AB+BC=AC}}}}[/tex]
[tex]\\[/tex]
We will find distance between points by distance formula-
[tex]\red{\underline{\boxed{\sf{Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}}}}[/tex]
[tex]\\(x_1,y_1) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (x_2,y_2)\\ \bull \frac{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }{} \bull \frac{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }{} \bull \\[/tex]
[tex]\begin{gathered}\\\\longrightarrow\quad \sf AB = \sqrt{(0 - 2) ^{2} + (3 - 4) ^{2} } \\\end{gathered} [/tex]
[tex]\begin{gathered}\\\\longrightarrow\quad \sf \sqrt{4 + 1 } \\\end{gathered} [/tex]
[tex]\begin{gathered}\\\\longrightarrow\quad \sf \sqrt{5 } \\\end{gathered} [/tex]
[tex]\\[/tex]
[tex]\begin{gathered}\\\\longrightarrow\quad \sf BC = \sqrt{(k - 0) ^{2} + (-4-3) ^{2} } \\\end{gathered} [/tex]
[tex]\begin{gathered}\\\\longrightarrow\quad \sf \sqrt{k^2+49 } \\\end{gathered}[/tex]
[tex]\\[/tex]
[tex]\begin{gathered}\\\\longrightarrow\quad \sf AC = \sqrt{(k - 2) ^{2} + (-4 - 4) ^{2} } \\\end{gathered} [/tex]
[tex]\begin{gathered}\\\\longrightarrow\quad \sf \sqrt{k^2+4-4k+64 } \\\end{gathered} [/tex]
[tex]\\[/tex]
[tex]\begin{gathered}\\\\longrightarrow\quad \sf AC = \sqrt{k^2-4k+68 } \\\end{gathered} [/tex]
[tex]\purple{ \underline { \boxed{ \sf{AB+BC=AC}}}}[/tex]
[tex]\\\underline{\tt\pink{Putting \:Values-}}\\[/tex]
[tex]\begin{gathered}\\\quad\longrightarrow\quad \sf \sqrt{5 }+\sqrt{k^2+49 } = \sqrt{k^2-4k+68 }[ \\\end{gathered} [/tex]
Squaring Both Sides-
[tex]\begin{gathered}\\\quad\longrightarrow\quad \sf (\sqrt{5 }+\sqrt{k^2+49 } )^2= (\sqrt{k^2-4k+68})^2 \\\end{gathered} [/tex]
