If a1 x + b1 y + c1 = 0 and a2 x + b2 y + c2 = 0 re such that a1 , b1

1.	If a1 x + b1 y + c1 = 0 and a2 x + b2 y + c2 = 0 re such that a1 , b1 , c1 a2 b2 and c2 are consecutive integers in the same order then find the values of x and y. A) -1, -2 B) 1, -2 C) -1, 2 D) 1, 2
Answer» Correct option is (B) 1, -2 Given that \(a_1,b_1,c_1,a_2,b_2\;and\;c_2\) are consecutive integers in the same order. Given system of equations is \( a_1 x + b_1 y + c _1= 0\) and \(a_2 x + b_2 y + c_2 = 0\) Then the solution of given system of equations by cross multiplication method : \(\frac x{b_1c_2-b_2c_1}=\frac y{c_1a_2-c_2a_1}=\frac1{a_1b_2-a_2b_1}\) \(\Rightarrow\) \(x=\frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}\;and\;y=\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}\) \(\Rightarrow x=\frac{(a_1+1)(a_1+5)-(a_1+4)(a_1+2)}{a_1(a_1+4)-(a_1+3)(a_1+1)}\) \(=\frac{(a_1^2+6a_1+5)-(a_1^2+6a_1+8)}{(a_1^2+4a_1)-(a_1^2+4a_1+3)}\) \(=\frac{5-8}{-3}\) \(=\frac{-3}{-3}=1\) and \(y=\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}\) \(=\frac{(a_1+2)(a_1+3)-(a_1+5)a_1}{a_1(a_1+4)-(a_1+3)(a_1+1)}\) \(=\frac{(a_1^2+5a_1+6)-(a_1^2+5a_1)}{(a_1^2+4a_1)-(a_1^2+4a_1+3)}\) \(=\frac6{-3}=-2\) Hence, the value of x and y are x = 1 & y = - 2. Correct option is B) 1, -2

If a1 x + b1 y + c1 = 0 and a2 x + b2 y + c2 = 0 re such that a1 , b1 , c1 a2 b2 and c2 are consecutive integers in the same order then find the values of x and y. A) -1, -2 B) 1, -2 C) -1, 2 D) 1, 2

Answer»

Correct option is (B) 1, -2

Given that \(a_1,b_1,c_1,a_2,b_2\;and\;c_2\) are consecutive integers in the same order.

Given system of equations is

\( a_1 x + b_1 y + c _1= 0\)

and \(a_2 x + b_2 y + c_2 = 0\)

Then the solution of given system of equations by cross multiplication method :

\(\frac x{b_1c_2-b_2c_1}=\frac y{c_1a_2-c_2a_1}=\frac1{a_1b_2-a_2b_1}\)

\(\Rightarrow\) \(x=\frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}\;and\;y=\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}\)

\(\Rightarrow x=\frac{(a_1+1)(a_1+5)-(a_1+4)(a_1+2)}{a_1(a_1+4)-(a_1+3)(a_1+1)}\)

\(=\frac{(a_1^2+6a_1+5)-(a_1^2+6a_1+8)}{(a_1^2+4a_1)-(a_1^2+4a_1+3)}\)

\(=\frac{5-8}{-3}\)

\(=\frac{-3}{-3}=1\)

and \(y=\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}\)

\(=\frac{(a_1+2)(a_1+3)-(a_1+5)a_1}{a_1(a_1+4)-(a_1+3)(a_1+1)}\)

\(=\frac{(a_1^2+5a_1+6)-(a_1^2+5a_1)}{(a_1^2+4a_1)-(a_1^2+4a_1+3)}\)

\(=\frac6{-3}=-2\)

Hence, the value of x and y are x = 1 & y = - 2.

Correct option is B) 1, -2