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  • Find the equation of one of the sides of an isosceles right angled triangle whose hypotenuse is given by 3x + 4y = 4 and the opposite vertex of the hypotenuse is (2, 2).

Find the equation of one of the sides of an isosceles right angled triangle whose hypotenuse is given by 3x + 4y = 4 and the opposite vertex of the hypotenuse is (2, 2).

Answers (1)

Equation of hypotenuse is 3x+4y=4 and opposite vertex is (2,2)

 Slope of the equation of hypotenuse is -\frac{3}{4}

 Now let the slope of AC be m \tan \theta =\left | \frac{\left ( m_{1}-m_{2} \right )}{1+m_{1}m_{2}} \right |

   Putting the values of m1and m2in the equation \tan 45^{0} =\left | \frac{\left ( m-\left ( -\frac{3}{4} \right ) \right )}{1-\frac{3}{4}m} \right |

 1=\left | \frac{m+\frac{3}{4}}{1-\frac{3}{4}m} \right |=\left | \frac{4m+3}{4-3m} \right |

 1=\pm \left | \frac{4m+3}{4-3m} \right |

\frac{4m+3}{4-3m} =1

 4m+3=4-3m     

     4m+3m=4-3              

   7m=1           m=\frac{1}{7}

  OR      -\frac{4m+3}{4-3m}=1       

 4m+3=-\left ( 4-3m \right )

  4m+3=-4+3m  

4m-3m=-4-3  

 m= -7

If m=1/7

equation of AC is y-y1=m(x-x1)   

y-2=1/7(x-2)

 7y-14=x-2  

 x-7y-2+14=0    

x-7y+12=0 

 If m= -7  then equation of AC is y-2=-(7)(x-2)  

 y-2=-7x+14    

7x+y=16 

The required equations are x-7y+12=0  and  7x+y=16

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