If the lines $a x - y - b = 0$ and $x - 2 b y + a = 0$ meet at the point (1, 2), then -

a = 1, b = 3

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a = 3, b = - 1

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a = 3, b = 1

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a = - 3, b = 1

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Solution

The correct option is C $a = 3, b = 1$
The given two lines meets at a point $(1, 2)$ means this point lies on both the lines.

i,e $x = 1 a n d y = 2$

$a (1) - 2 - b = 0$

$. a - b = 2$ .... equation 1

And

$(1) - 2 (b) (2) + a = 0$

$a - 4 b = - 1$ ....... equation 2

Subtract equation 2 from 1

$3 b = 3$

$b = 1$

Now put value of b in equation 1

we get $a = 3$

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Similar questions

If the lines a_{1} x + b_{1} y + 1 = 0, a_{2} x + b_{2} y + 1 = 0 and a_{3} x + b_{3} y + 1 = 0 are concurrent, then the points (a_{1}, b_{1}), (a_{2}, b_{2}) and (a_{3}, b_{3}) will be collinear .

Q. If

A = ∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} & c_{1} a_{2} & b_{2} & c_{2} a_{3} & b_{3} & a_{3} \end{matrix} ∣ ∣ ∣ ∣ \neq 0,

then the system of equations

a_{1} x + b_{1} y + c_{1} z = 0, a_{2} x + b_{2} y + c_{2} z = 0

and

a_{3} x + b_{3} y + c_{3} z = 0

has:

Q. Equation of the common tangent to the parabola

y^{2} = 4 a x

and

x^{2} = 4 b y

Three lines given by $L_{1} : a_{1} x + b_{1} y + c_{1} = 0, L_{2} : a_{2} x + b_{2} y + c_{2} = 0 and L_{3} : a_{3} x + b_{3} y + c_{3} = 0$ are always concurrent given that

$∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} & c_{1} a_{2} & b_{2} & c_{2} a_{3} & b_{3} & c_{3} \end{matrix} ∣ ∣ ∣ ∣ = 0$

Q. Three lines given by

L_{1} : a_{1} x + b_{1} y + c = 0, L_{2} : a_{2} x + b_{2} y + c_{2} = 0 a n d L_{3} : a_{3} x + b_{3} y + c_{3} = 0

and given that

∣ ∣ ∣ ∣ \begin{matrix} a_{1} & b_{1} & c_{1} a_{2} & b_{2} & c_{2} a_{3} & b_{3} & c_{3} \end{matrix} ∣ ∣ ∣ ∣

=0, then these lines

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If the lines ax−y−b=0 and x−2by+a=0 meet at the point (1, 2), then -

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