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Condition for Two Lines to Be Parallel

What will be ...

Question

What will be the solution of these equations $a x + b y = a - b$ , $b x - a y = a + b$

x = 1, y = 2

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x = 2, y = - 1

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x = - 2, y = - 2

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x = 1, y = - 1

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Solution

The correct option is C $x = 1, y = - 1$
$a x + b y = a - b$ .......... (1)
$b x - a y = a + b$ .......... (2)

Multiply equation (1) with $a$ and equation (2) with $b$ ,

$a^{2} x + a b y = a^{2} - a b$ ..........(3)
$b^{2} x - a b y = a b + b^{2}$ ......... (4)

Adding (3) and (4), we get,

$(a^{2} + b^{2}) x = a^{2} + b^{2}$

$\Rightarrow x = 1$

Now, from (1),

$a x + b y = a - b$

$a + b y = a - b$

$b y = - b$

$y = - 1$

Suggest Corrections

Similar questions

x

y

a x + b y = 1

b x + a y = \frac{(a + b)^{2}}{a^{2} + b^{2}} - 1

b x + a y = \frac{2 a b}{a^{2} + b^{2}}

Factorise: $2 a x + 2 b x - a y - b y$

Column I	Column II
(a) Solution of $a x + b y = a - b, b x - a y = a + b is . . . . . . .$	(p) $x = 2, y = - 2$
(b) Solution of $3 x - 2 y = 10, 5 x + 3 y = 4 is . . . . . . .$	(q) $x = a, y = b$
(c) Solution of $2 x - y = 5, 3 x + 2 y = 11 is . . . . . . .$	(r) $x = 1, y = - 1$
(d) Solution of $x + y = a + b, a x - b y = a^{2} - b^{2} is . . . . . . .$	(s) $x = 3, y = 1$

(i) (2 a - 1) x - 3 y = 5 3 x + (b - 2) y = 3 (i i) 2 x - (2 a + 5) y = 5 (2 b + 1) x - 9 y = 15 (i i i) (a - 1) x + 3 y = 2 6 x + (1 - 2 b) y = 6 (i v) 3 x + 4 y = 12 (a + b) x + 2 (a - b) y = 5 a - 1 (v) 2 x + 3 y = 7 (a - b) x + (a + b) y = 3 a + b - 2 (v i) 2 x + 3 y - 7 = 0 (a - 1) x + (a + 1) y = (3 a - 1) (v i i) 2 x + 3 y = 7 (a - 1) x + (a + 2) y = 3 a

(v i i i) x + 2 y = 1 (a - b) x + (a + b) y = a + b - 2

(i x) 2 x + 3 y = 7 2 a x + a y = 28 - b y

x, y, z

x + y + z = 0

(1 + a) x + (2 + a)

y - 8 z

= 0, x - (1 + a) y + (2 + a) z = 0

a =

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