For what value of k will the system of linear equations KX 3y 3 and 12x Ky 6 represent parallel lines?

The given system of equations:kx + 3y = 3kx + 3y - 3 = 0                    ….(i)12x + ky = 612x + ky - 6 = 0                    ….(ii)

These equations are of the following form:

`a_1x+b_1y+c_1 = 0, a_2x+b_2y+c_2 = 0`where, `a_1 = k, b_1= 3, c_1= -3 and a_2 = 12, b_2 = k, c_2= –6`In order that the given system has no solution, we must have:`(a_1)/(a_2) = (b_1)/(b_2) ≠ (c_1)/(c_2)``i .e.,  k/12 = 3/k ≠ (−3)/(−6)``k/12 = 3/k and 3/k ≠ 1/2``⇒ k^2 = 36 and k ≠ 6``⇒ k = ±6 and k ≠ 6`

Hence, the given system of equations has no solution when k is equal to -6.

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The given system of equations:3x - y - 5 = 0                      ….(i)And, 6x - 2y + k = 0             ….(ii)These equations are of the following form:`a_1x+b_1y+c_1 = 0, a_2x+b_2y+c_2 = 0`where,`a_1 = 3, b_1= -1, c_1= -5 and a_2 = 6, b_2= -2, c_2 = k`In order that the given system has no solution, we must have:`(a_1)/(a_2) = (b_1)/(b_2) ≠ (c_1)/(c_2)`` i.e., 3/6 = (−1)/(−2) ≠ −5/k``⇒(−1)/(−2) ≠ (−5)/k ⇒ k ≠ -10`

Hence, equations (i) and (ii) will have no solution if k ≠ -10.

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The given system of equations can be written askx + 3y + 3 - k = 0                       ….(i)12x + ky - k = 0                           ….(ii)This system of the form:`a_1x+b_1y+c_1 = 0``a_2x+b_2y+c_2 = 0`

where, `a_1 = k, b_1= 3, c_1 = 3 - k and a_2 = 12, b_2 = k, c_2= –k`

For the given system of linear equations to have no solution, we must have:`(a_1)/(a_2) = (b_1)/(b_2) ≠ (c_1)/(c_2)``⇒ k/12 = 3/k ≠ (3−k)/(−k)``⇒k/12 = 3/k and 3/k ≠ (3−k)/(−k)``⇒ k^2 = 36 and -3 ≠ 3 - k`⇒ k = ±6 and k ≠ 6⇒k = -6

Hence, k = -6.

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The given system of equations:5x - 3y = 0                 ….(i)2x + ky = 0                 ….(ii)These equations are of the following form:`a_1x+b_1y+c_1 = 0, a_2x+b_2y+c_2 = 0`where, `a_1 = 5, b_1= -3, c_1 = 0 and a_2 = 2, b_2 = k, c_2 = 0`For a non-zero solution, we must have:`(a_1)/(a_2) = (b_1)/(b_2)``⇒ 5/2 = (−3)/k``⇒5k = -6 ⇒ k = (−6)/5`

Hence, the required value of k is `(−6)/5`.

Question 30 Pair of Linear Equations in Two Variables - Exercise 3.3

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Answer:

The given pair of linear equations is

kx + 3y = k – 3 …(i)

12x + ky = k …(ii)

On comparing the equations (i) and (ii) with ax + by = c = 0,

We get,

a1 = k, b1 = 3, c1 = -(k – 3)

a2 = 12, b2 = k, c2 = – k

Then,

a1 /a2 = k/12

b1 /b2 = 3/k

c1 /c2 = (k-3)/k

For no solution of the pair of linear equations,

a1/a2 = b1/b2≠ c1/c2

k/12 = 3/k ≠ (k-3)/k

Taking first two parts, we get

k/12 = 3/k

k2 = 36

k = + 6

Taking last two parts, we get

3/k ≠ (k-3)/k

3k ≠ k(k – 3)

k2 – 6k ≠ 0

so, k ≠ 0,6

Therefore, value of k for which the given pair of linear equations has no solution is k = – 6.

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