InterviewSolution
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InterviewSolution
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Draw the graph of the equation 3x-4y =12. comment on: (i)x=0, y=3 is a solution of the equation. (ii) The abscissa i.e. the value of x can never be 100 units. (iii) Sum of intercept (parts made by straight line on the axes) on the axed is 7 units. (iv) Length of line segment between the axes is 5 units (v) Area of triangle formed by the line 3x-4y =12 and co ordinate axes. |
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Answer» Solution :We have 3x-4y =12 3x=4y+12 x=4y+12 3 Let y =0 `rArr` x=4 Let y = -3 `rArr` x= 0 Table for above values  Plot the points A(4,0) and B(0,-3) on the graph paper and join them . producer this line on both sides (i)we see form the graph , that (0,3) does not lie on the given equation . so x=0 , y=3 is not the solution of 3x-4y=12 Also we can check algebraically (without seeing the graph) putting x=0 and y =3 in 3x-4y=12 we get 3(0) - (3) =12 `rArr` -12=12 which is not true. Spo , x =0,y =3 connot be the solution of given eqautio 3x-4y = 12 (ii) if we take any real value of x, then we get the corresponding value of y and vice versa from the equation 3x-4y =12 so x= 100 can also be taken (iii) From the graph we see the intercept made by line on the y axis =-3 so sum of intercepts =4+(-3)=1 There , it is not true that sunm of inteceps is 7. (iv) Since length of OA =4 units and length of OB =3 units `therefore` In RIGHT angle triangle AOB by pythagoras stheorem `AB^(2)=OA^(2)+OB^(2)` `=(4)^(2)+(3)^(2)=16+9=25` `AB=sqrt(25)=5 units` So length of line SEGMENT AB =5 units (v) Area of triangle =`(1)/(2)xxOAxxOB=(1)/(2)xx4xx3` units =6 sq units 
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