Question

For what value of k are the points A(8,1),B(3,2k) and C(k,5) collinear

Answer 1

${\underline{\sf{Question}}}$

For what value of k are the points A(8,1),B(3,2k) and C(k,5) collinear.

${\underline{\sf{Theory }}}$

Area of a triangle :

The area of a triangle, the coordinates of whose vertices are $\sf\:(x_{1},y_{1}),(x_{2},y_{2}),\:and\:(x_{3},y_{3})$.

Area of ∆ $\sf=\dfrac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$

or

Area of ∆=$\sf\dfrac{1}{2}$$\left|\begin{array}{ccc}x_{1}&y_{1}&1\\x_{2}&y_{2}&1\\x_{3}&y_{3}&1\end{array}\right|$

⇒If three points are collinear then

Area of ∆ =0

${\underline{\sf{Answer. }}}$

Let the points be A (8,1), B (3,2k)band C (k,5). Then ,

$\sf\:x_{1}=8,x_{2}=3\:and\:x_{3}=k..(1)$

$\sf\:y_{1}=1,y_{2}=2k\:and\:y_{3}=5. ..(2)$

If the given points are collinear , then

Area ∆ABC = 0

$\sf\dfrac{1}{2}$$\left|\begin{array}{ccc}x_{1}&y_{1}&1\\x_{2}&y_{2}&1\\x_{3}&y_{3}&1\end{array}\right|$=0

Put the values of equation (1) and (2)

$\sf\dfrac{1}{2}$$\left|\begin{array}{ccc}8&1&1\\3&2k&1\\k&5&1\end{array}\right|$= 0

⇒$\left|\begin{array}{ccc}8&1&1\\3&2k&1\\k&5&1\end{array}\right|$= 0

⇒8[2k-5]-(1)[3-k]+1[15-2k²]= 0

⇒2k²-17k+28=0

By Quadratic Formula :

$\sf{x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}}$

$\implies\sf{k=\dfrac{17\pm\sqrt{(-17)^2-4(2)(28)}}{2\times2}}$

$\implies\sf{k=\dfrac{17\pm\sqrt{289-224}}{4}}$

$\implies\sf{k=\dfrac{17\pm\sqrt{65}}{4}}$

$\implies\sf{k=\dfrac{17}{4}\pm\dfrac{\sqrt{65}}{4}}$

Therefore,the value of k $\implies\sf{=\dfrac{17}{4}\pm\dfrac{\sqrt{65}}{4}}$

Answer 2

Area of a triangle :

The area of a triangle, the coordinates of whose vertices are \sf\:(x_{1},y_{1}),(x_{2},y_{2}),\:and\:(x_{3},y_{3})(x

1

,y

1

),(x

2

,y

2

),and(x

3

,y

3

) .

Area of ∆ \sf=\dfrac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]=

2

1

[x

1

(y

2

−y

3

)+x

2

(y

3

−y

1

)+x

3

(y

1

−y

2

)]

or

Area of ∆=\sf\dfrac{1}{2}

2

1

\begin{gathered}\left|\begin{array}{ccc}x_{1}&y_{1}&1\\x_{2}&y_{2}&1\\x_{3}&y_{3}&1\end{array}\right|\end{gathered}

∣

∣

∣

∣

∣

∣

∣

x

1

x

2

x

3

y

1

y

2

y

3

1

1

1

∣

∣

∣

∣

∣

∣

∣

⇒If three points are collinear then

Area of ∆ =0

{\underline{\sf{Answer. }}}

Answer.

Let the points be A (8,1), B (3,2k)band C (k,5). Then ,

\sf\:x_{1}=8,x_{2}=3\:and\:x_{3}=k..(1)x

1

=8,x

2

=3andx

3

=8,x

2

=3andx

3

=k..(1)

\sf\:y_{1}=1,y_{2}=2k\:and\:y_{3}=5. ..(2)y

1

=1,y

2

=2kandy

3

=5...(2)

If the given points are collinear , then

Area ∆ABC = 0

\sf\dfrac{1}{2}

2

1

\begin{gathered}\left|\begin{array}{ccc}x_{1}&y_{1}&1\\x_{2}&y_{2}&1\\x_{3}&y_{3}&1\end{array}\right|\end{gathered}

∣

∣

∣

∣

∣

∣

∣

x

1

x

2

x

3

y

1

y

2

y

3

1

1

1

∣

∣

∣

∣

∣

∣

∣

=0

Put the values of equation (1) and (2)

\sf\dfrac{1}{2}

2

1

\begin{gathered}\left|\begin{array}{ccc}8&1&1\\3&2k&1\\k&5&1\end{array}\right|\end{gathered}

∣

∣

∣

∣

∣

∣

∣

8

3

k

1

2k

5

1

1

1

∣

∣

∣

∣

∣

∣

∣

= 0

⇒\begin{gathered}\left|\begin{array}{ccc}8&1&1\\3&2k&1\\k&5&1\end{array}\right|\end{gathered}

∣

∣

∣

∣

∣

∣

∣

8

3

k

1

2k

5

1

1

1

∣

∣

∣

∣

∣

∣

∣

= 0

⇒8[2k-5]-(1)[3-k]+1[15-2k²]= 0

⇒2k²-17k+28=0

By Quadratic Formula :

\sf{x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}}x=

2a

−b±

b

2

−4ac

\implies\sf{k=\dfrac{17\pm\sqrt{(-17)^2-4(2)(28)}}{2\times2}}⟹k=

2×2

17±

(−17)

2

−4(2)(28)

\implies\sf{k=\dfrac{17\pm\sqrt{289-224}}{4}}⟹k=

4

17±

289−224

\implies\sf{k=\dfrac{17\pm\sqrt{65}}{4}}⟹k=

4

17±

65

\implies\sf{k=\dfrac{17}{4}\pm\dfrac{\sqrt{65}}{4}}⟹k=

4

17

±

4

65

Therefore,the value of k \implies\sf{=\dfrac{17}{4}\pm\dfrac{\sqrt{65}}{4}}⟹=

4

17

±

4

=k..(1)