Question

2. Find the value of 'K'for which the points are collinear

(7,-2) (5, 1) (3, K)​

Answer 1

Question :

Find the value of 'K'for which the points are collinear (7,-2) (5, 1) (3, K)

Theory :

Area of the ∆ formed by three points

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

⇒If three points are collinear then

Area of ∆ formed by these points =0

Solution :

Let the poitns be A (7,-2), B (5, 1)band C (3, K). Then ,

$\sf\:x_{1}=7,x_{2}=5\:and\:x_{3}=3..(1)$

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

If the given points are collinear , then

Area ∆ABC = 0

Area of the ∆ABC = 0

$\sf\:\implies=\dfrac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$

Now put the values of equation (1)&(2)

$\sf \implies \:\frac{1}{2}[7(1 - k) + 5(k - ( - 2)) + 3( - 2 - 1)]= 0$$\sf \implies7(1 - k) + 5(k + 2) + 3( - 3) = 0$

$\sf \implies7 - 7k + 5k + 10 - 9 = 0$

$\sf \implies10 - 2 = 7k - 5k$

$\sf \implies2k = 8$

$\sf \implies \: k = \frac{ \cancel{8}}{ \cancel{2}}$

$\sf \implies \: k = 4$

Therefore,the value of k = 4

Answer 2

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

Find the value of 'K'for which the points are collinear (7,-2) (5, 1) (3, K)

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

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

Solution :

Let the poitns be A (7,-2), B (5, 1)band C (3, K). Then ,

$\sf\:x_{1}=7,x_{2}=5\:and\:x_{3}=3..(1)$

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

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}7&-2&1\\5&1&1\\3&k&1\end{array}\right|$= 0

⇒$\left|\begin{array}{ccc}7&-2&1\\5&1&1\\3&k&1\end{array}\right|$= 0

⇒7[1-k]-(-2)[5-3]+1[5k-3]= 0

⇒7-7k +10 -6+5k-3= 0

⇒ 17-9= 7k-5k

⇒2k = 8

⇒k = 4

Therefore,the value of k = 4