Question

If $\frac { \mathrm { x } } { l }$ + = 1 be any line passing through the intersection point of the lines = 1 and = 1 then-

• A. $\frac { 1 } { \mathrm { a } } + \frac { 1 } { \mathrm {~m} } = \frac { 1 } { \mathrm {~b} } + \frac { 1 } { l }$
• B.
• C. $\frac { 1 } { l } + \frac { 1 } { \mathrm {~m} } = \frac { 1 } { \mathrm { a } } + \frac { 1 } { \mathrm {~b} }$
• D. None of these

Answer 1

Answer: (C)

$\frac { 1 } { l } + \frac { 1 } { \mathrm {~m} } = \frac { 1 } { \mathrm { a } } + \frac { 1 } { \mathrm {~b} }$

Answer 2

Equation of any line through the intersection point of the given lines must be of the type

$\left( \frac { \mathrm { x } } { \mathrm { a } } + \frac { \mathrm { y } } { \mathrm { b } } - 1 \right)$ + l $\left( \frac { \mathrm { x } } { \mathrm { b } } + \frac { \mathrm { y } } { \mathrm { a } } - 1 \right)$ = 0

i.e. x $\left( \frac { 1 } { a } + \frac { \lambda } { b } \right)$ + y $\left( \frac { 1 } { b } + \frac { \lambda } { a } \right)$ – (1 + l) = 0

Comparing it with the given equation, we have

$\frac { 1 } { l }$ = $\frac { 1 } { 1 + \lambda }$ … (1)

and … (2)

Thus, we have$\frac { \mathrm { ab } } { \mathrm { m } }$ =

Adding, we have ab $\left( \frac { 1 } { l } + \frac { 1 } { \mathrm {~m} } \right)$ =

i.e. $\frac { 1 } { l } + \frac { 1 } { \mathrm {~m} } = \frac { 1 } { \mathrm { a } } + \frac { 1 } { \mathrm {~b} }$.