Question

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

Answer 1

since (1,2) is the point of

intersection of above

mentioned lines ...

hence it will satisfy the

equations is the lines

putting x = 1 and y = 2 in the

1st equation we have

a(1)–(2)–b= 0

hence

a – b = 2 ....... 3

and put x => 1 and y => 2 in

the 2nd equation we have

(1)–2b(2)+a=0

a–4b = –1 ...........4

from 3rd equation b=> a–2,,

put it in 4th equation

a –4(a–2)=0

–3a+8=–1

a = 9/3 = 3

and b=> 3–2

=> 1

a => 3 and b=> 1

Answer 2

Answer :

As given that ax - y - b = 0 and x - 2by + a = 0 meet at point (1 , 2). Therefore it is clear that here the value of x is 1 and value of y is 2.

So, To find the value of a and b, we need to put the given values of x and y in the given equations.

we get ---
$\binom{a(1) - 2 - b = 0}{1 - 2b(2) + a =0 } \\ \\ firstly \: solve \: the \: equation \: for \: a \: \\ \\ \binom{a - 2 - b = 0}{1 - 2b \times 2 + a = 0} \\ \\ \binom{a = 2 + b}{1 - 2b \times 2 + a = 0}$

$substitute \: the \: given \: value \: of \: a \: \\ into \: the \: equation \: 1 - 2b \times 2 + 2 + b = 0 \: \: we \: get \\ \\ 1 - 2b \times 2 + 2 + b = 0 \\ \\ solve \: the \: equation \: for \: b \\ \\ 1 - 4b + 2 + b = 0 \\ \\ 3 - 3b = 0 \\ \\ - 3b = - 3 \\ \\ b = \frac{3}{3} \\ \\ b = 1$

$substitute \: the \: given \: value \: of \: b \: into \:the \: equation \\ a = 2 + b \: \: we \: get \\ \\ a = 2 + 1 \\ \\ a = 3 \\ \\ therefore \\ \\ a = 3 \: \: and \: \: b \: = 1$