Question

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

Answer 1

Step-by-step explanation:

k=-3/2

k=5

hope it is helpful

Answer 2

⇒SOLUTION:-

We know that ,Area of triangle

$= \displaystyle \sf \dfrac{1}{2} [x _{1}(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

Here,

$\sf x_1=3 \: \: \: \: \: y_1=1$

$\sf x_2=3 \: \: \: \: y_2=-2k$

$\sf x_3=k \: \: \: \: y_3=-5$

Area of triangle

⠀⠀⠀⠀

$\implies \sf \dfrac{1}{2} [8(-2k+5)+3(-5-1)+k(1+2k)]$

⠀⠀⠀⠀

$\implies \sf \dfrac{1}{2} [-16k+40+3(-6)+k+2k^{2}]$

⠀⠀⠀⠀

$\implies \sf \dfrac{1}{2} [-16k+40-18+k+2k^{2}]$

⠀⠀⠀⠀

$\implies \sf \dfrac{1}{2} [2k^{2}-15k+22]$

•Given points will be collinear ,if area of triangle formed by them is zero

⠀⠀⠀⠀

$\therefore \sf\dfrac{1}{2} [2k^{2}-15k+22]=0$

⠀⠀⠀⠀

$\implies \sf {2k}^{2} - 16k + 22 = 0$

⠀⠀⠀⠀

$\implies \sf {2k}^{2} - 4k - 11k + 22 = 0$

⠀⠀⠀⠀

$\implies \sf2k(k - 2) - 11(k - 2) = 0$

⠀⠀⠀⠀

$\implies \sf(2k - 11)(k - 2) = 0$

⠀⠀⠀⠀

$\implies \sf2k - 11 = 0 \: \: or \: \: k - 2 = 0$

⠀⠀⠀⠀

$\implies \sf k = \dfrac{11}{2} \: \: or \: \: k = 2$

⠀⠀⠀⠀

→Hence,the given points are collinear for k=2 or k=11/2