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  • Can someone explain The equation of the straight line passing through the point (4, 3) and making intercepts on the co-ordinate axes whose sum is –1 is

The equation of the straight line passing through the point (4, 3) and making intercepts on the co-ordinate axes whose sum is –1 is

  • Option 1)

    \frac{x}{2}+\frac{y}{3}=1\; and\; \frac{x}{2}+\frac{y}{1}=1\;

  • Option 2)

    \frac{x}{2}-\frac{y}{3}=-1\; and\; \frac{x}{-2}+\frac{y}{1}=1\;

  • Option 3)

    \frac{x}{2}+\frac{y}{3}=-1\; and\; \frac{x}{-2}+\frac{y}{1}=-1\;

  • Option 4)

    \frac{x}{2}-\frac{y}{3}=1\; and\; \frac{x}{-2}+\frac{y}{1}=1\;

 

Answers (1)

As we learnt in

Intercept form of a straight line -

\frac{x}{a}+\frac{y}{b}=1

 

- wherein

a and b are the x-intercept and y -intercept respectively.

 

 Let the equation of straight line be 

\frac{x}{a}+\frac{y}{b}=1

i.e.  \frac{4}{a}+\frac{3}{b}=1                                                ..........(1)

also\ a+b=-1                                         ...........(2)

4b+3a=ab

4(-1-a)+3a=a(-1-a)

-4 - a = - a - a2

\Rightarrow a^{2}=4\, \, \, \, \Rightarrow a=\pm2

and a+b=-1

if a=2, b=-3

a=-2, b=1 


Option 1)

\frac{x}{2}+\frac{y}{3}=1\; and\; \frac{x}{2}+\frac{y}{1}=1\;

This solution is incorrect 

Option 2)

\frac{x}{2}-\frac{y}{3}=-1\; and\; \frac{x}{-2}+\frac{y}{1}=1\;

This solution is incorrect 

Option 3)

\frac{x}{2}+\frac{y}{3}=-1\; and\; \frac{x}{-2}+\frac{y}{1}=-1\;

This solution is incorrect 

Option 4)

\frac{x}{2}-\frac{y}{3}=1\; and\; \frac{x}{-2}+\frac{y}{1}=1\;

This solution is correct 

Posted by

Vakul

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