Find the Equation of the line AB when it meets x-axis at (-3,0) with slope = -2

Find the Equation of the line AB when it meets x-axis at (-3,0) with slope = -2


Sol :

Here,

                m = -2

                x₁ = -3  and y₁ = 0

Therefore           Equation of the line AB:

                                Y-0 = -2(x+3)

Or                           Y = -2x-3

Or                           2x+y+6 = 0

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Find the Equation of the line which is passing through the point (0,0) with slope m.

Sol :

Here

              X₁ = 0 and y₁ = 0

             Slope = m

Therefore,           Equation of the line is :

                            

                             (y-0) = m(x-0)

Or                          y = mx

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Find the Equation of the line which is passing through (-4,3) with slope 1/2

Find the Equation of the line which is passing through (-4,3) with slope 1/2


Sol :

We know that Equation of line will be (y-y₁) = m(x-x₁) when slope of line and a point is given.

Here 
    

x₁ = -4 ,  y₁ = 3  and  m = 1/2

 Therefore,  The Required equation of the line is 

                      (y-3) = 1/2(x+4).

Or,                2y-6 = x+4

Or,         x-2y+10 = 0

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Write the equations for x and y-axis.

Write the equations for x and y-axis.

Sol :

Equation of x-axis is y = 0.

and

Equation of y-axis is x = 0.

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If three points (h.0),(a,b) and (0,k) lie on line, show that a/h+b/k=1

Q.No. 13 of Exercise 10.1 of Class 11th

If three points (h.0),(a,b) and (0,k) lie on line, show that a/h+b/k=1.

Sol :
Let, 
   The given points are A(h,0), B(a,b), C(0,k)
   These points lie on the same line

Therefore,  
      Slope of AB = Slope of BC.


Here, 
Slope of AB = b-0/a-h = b/a-h

Slope of BC = k-b/0-a = k-b/-a


Now

              b/a-h = k-b/-a
Or            -ab = (a-h)(k-b)               [By Cross Multiplication]
Or            -ab = ak-ab-hk+bh
Or               0 = ak-hk+bh
Or     ak+bh  = hk

Dividing both side by hk, we get
            
                        ak/hk + bh/hk = hk/hk
Or                   a/h + b/k = 1

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The Slope of a line is double of the slope of another line. If tangent of the angle between them is 1/3, find the slopes of the line.

Q.No. 11 of Exercise 10.1 of Class 11th

The angle between x-axis and the line joining the points A(3,-1) and (4,-2).

Sol :

Let m₁ and m_{2 be the slopes of the two lines.}

_{Given that}        m₁ = 2m₂

 
Let x be the angle between the lines.

We know that,         tanx = m₁ – m₂/1+m₁m₂

                

_{Therefore,               1/3   =}   m₁ – m₂/1+m₁m₂

_Putting       m₁ = 2m_2,

_{Therefore,              1/3  =} 2m₂ – m₂/1+2m₂m₂ 

                                 

      _{Or                          1/3   =}   m₂/1+2m₂²

       ^Or                          3m₂  = 1+2m₂²

        Or                2m₂²-3m₂+1 = 0

       Or          2m₂²-2m₂- m₂+1 = 0

      Or        2m₂(m₂-1)-1(m₂-1) = 0

      Or              (2m₂-1)( m₂-1) = 0

Therefore,        m_{2 = 1/2      OR}          m_{2 = 1}   

Now,

         For   m_{2 = 1/2}         m_{1 = 1}

_{and   For}     m_{2 = 1}          m_{1= 2}

_{Hence, Slope of these lines are  1, 1/2 or 2,1.} 

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