Question

if the lines 2x + 3y = 6 and 6x-ky = 18 are coincident lines, then find the value of k.​

Answer 1

SOLUTION

GIVEN

The lines 2x + 3y = 6 and 6x - ky = 18 are coincident lines

TO DETERMINE

The value of k

CONCEPT TO BE IMPLEMENTED

A pair of Straight Lines

$\displaystyle \sf{ a_1x+b_1y+c_1=0 \: \: and \: \: \: a_2x+b_2y+c_2=0}$

is said to be coincident lines if

$\displaystyle \sf{ \: \frac{a_1}{a_2} = \frac{b_1}{b_2} = \: \frac{c_1}{c_2}}$

EVALUATION

Here the given lines are

2x + 3y = 6 and 6x - ky = 18

Now the lines are coincident lines if

$\displaystyle\sf{ \frac{2}{6} = \frac{3}{ - k} = \frac{6}{18} }$

$\displaystyle\sf{ \implies \: \frac{1}{3} = \frac{3}{ - k} = \frac{1}{3} }$

$\displaystyle\sf{ \implies \: \frac{3}{ - k} = \frac{1}{3} }$

$\displaystyle\sf{ \implies \: - k = 9 }$

$\displaystyle\sf{ \implies \: k = - 9 }$

FINAL ANSWER

Hence the required value of k = - 9

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Answer 2

Given :  lines 2x + 3y = 6 and 6x-ky = 18 are coincident lines

To Find :   the value of k.​

Solution:

Pair of linear equations

a₁x  +  b₁y + c₁  =  0

a₂x  +  b₂y + c₂  =  0

Consistent

if a₁/a₂ ≠ b₁/b₂   (unique solution  and lines intersects each others)

  a₁/a₂ = b₁/b₂ = c₁/c₂   (infinite solutions and line coincide each other )

Inconsistent

if  a₁/a₂ = b₁/b₂ ≠  c₁/c₂  ( No solution , lines are parallel to each other)

2x + 3y = 6

6x-ky = 18

  a₁/a₂ = b₁/b₂ = c₁/c₂  coincident lines

=> 2/6  =  3/-k  = 6/18

=> 1/3 = 3/-k  = 1/3

=> k =  - 9

value of k is - 9

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