Question

The value of 'K' for which the equations 3x-y+8=0 and 6x+ Ky= -16 represent coincident lines, is:​

Answer 1

follow me....

.............

Answer 2

$\pink{\underline{\sf{Given: }}}$

Two equations,

3x - y + 8 = 0

6x + Ky = -16

$\blue{\underline{\sf{So,}}}$

3x - y + 8 = 0 ......$\red{(1)}$

6x + Ky + 16 = 0 .....$\red{(2)}$

$\green{\underline{\sf{Now,\: to\:\:be\;\:found-}}}$

The value of 'K' for which the equations $\red{(1)}$ and $\red{(2)}$ represent coincident lines.

$\blue{\underline{\tt{Here \:\:is\:\: more\:\: information -}}}$

$\begin{array}{| c | c | c | c |}\cline{1-4} \bf Compara& \bf Graphical & \bf Algebraic & \bf Consistency\\ \cline{4-4} \bf the\:\:ratios&\bf representation&\bf interpretation&-\\ \cline{1-4}\bf\dfrac{a_{1} }{b_{1} } \neq \dfrac{ b_{1}}{b_{2}} & \sf Intersecting\:\:lines & \sf One\:\:Solution(Unique)& \sf Consistent \\ \cline{1-4} \bf \dfrac{a_{1} }{b_{1} }=\dfrac{ b_{1}}{b_{2}}=\dfrac{c_{1}}{c_{2}}& \sf Coincident\:\:lines& \sf Infinity\:\:many\:\:solutions& \sf Consistent(dependent)\\ \cline{1-4} \bf \dfrac{a_{1} }{b_{1} }=\dfrac{ b_{1}}{b_{2}}\neq\dfrac{c_{1}}{c_{2}}&\sf Parallel\:\:lines& \bf No\:\:Solution& \sf Inconsistent\\ \cline{1-4} \end{array}$

Now,

For coincident or infinity many solution we have to prove,

$\boxed{\boxed{\bf \dfrac{a_{1} }{b_{1} }=\dfrac{ b_{1}}{b_{2}}=\dfrac{c_{1}}{c_{2}} }}$

So,

a₁ = 3, b₁ = (-1), c₁ = 8

And

a₂ = 6, b₂ = K, c₂ = 16

so,

$\bf \dfrac{a_{1} }{b_{1} }=\dfrac{ b_{1}}{b_{2}}=\dfrac{c_{1}}{c_{2}}$

$\implies \bf \dfrac{3 }{6 }=\dfrac{-1}{K}=\dfrac{8}{16}$

$\implies \bf \dfrac{1 }{2 }=\dfrac{-1}{K}=\dfrac{1}{2}$

$\implies \bf \dfrac{1 }{2 }=\dfrac{-1}{K}$

$\implies \boxed{\boxed{\bf K = (-2)}}$

Hence,

For the the value $\red{\boxed{-2}}$ = K the given equations will have coincident lines.