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34. Find the equation of the straight line which
.
passes through the point (3,1) and makes
intercepts on the axes the difference of
whose length is 4.


Sagot :

Answer:

The equation of the straight line passing through the point ( 3,1 )

                      [tex]\frac{x}{2} + \frac{y}{-2} = 1[/tex]

Step-by-step explanation:

Step(i):-

The equation of the straight line passing through the point ( 3,1 )

                      [tex]\frac{x}{a} + \frac{y}{b} = 1[/tex]

                     [tex]\frac{3}{a} + \frac{1}{b} = 1[/tex]

                    3b + a = ab ...(i)

Given the difference of length is  4

                      a-b = 4

                       b =  a - 4 ...(ii)

Step(ii):-

substitute b=a-4 in equation (i) , we get

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

               3a - 12+ a =  - 4 a + a²  

                 a² - 8 a + 12 =0

Find the factors of 'a'

                 a² - 6a -2a +12 =0

                a (a-6) -2(a-6) =0

             a =2 and a=6

                  we know that a-b =4

             put       a = 2

                          2 - b =4

                            b = -2

The equation of the straight line whose intercepts on the axes

                    [tex]\frac{x}{a} + \frac{y}{b} = 1[/tex]

                   [tex]\frac{x}{2} + \frac{y}{-2} = 1[/tex]

The equation of the straight line

                      [tex]\frac{x}{2} + \frac{y}{-2} = 1[/tex]

                   

Verification:-

The equation of the straight line passing through the point (3,1)

                      [tex]\frac{x}{2} + \frac{y}{-2} = 1[/tex]

Put x =3 and y=1

                  [tex]\frac{3}{2} + \frac{1}{-2} = 1\\\frac{2}{2} =1\\[/tex]

                 1 = 1

∴  The point (3,1) is satisfies the equation