SHOW THAT THE THREE POINTS (5,1), (1,-1) AND (11,4) LIE ON A STRAIGHT LINE. FUTHER FIND:
(i) ITS INTERCEPTS ON THE AXES (ANS IN THE TB : 3, -3/2)
(ii) THE LENGTH OF THE PORTION OF THE LINE INTERCEPTED BETWEEN THE AXES (ANS IN TB: 3√5/2)
(iii) THE SLOPE OF LINE ( ANS IN TB: 1/2)

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Please find below the solution to the asked query:

We have point A5,1 and B1,-1 and C11,4Equation of line AB will be given by:y-1=-1-11-5x-5y-1=12x-52y-2=x-5x-2y-5+2=0x-2y-3=0Now putting point C in R.H.S. we get:11-8-3=8-8=0Hence C lies on line AB.Hence points A5,1 and B1,-1 and C11,4 lie on a straight line.Againx-2y-3=0Slope of line=-Coefficient of xCoefficient of y=-1-2=12 Answerx-2y-3=0x-2y=3x3-2y3=1x3+y-32=1On comparing it with intercept form xa+yb=1X-axis intercepta=3 AnswerY-axis interceptb=-32 AnswerLength intercepted between axes=a2+b2=32+-322=32+324=31+14=354=325 units Answer

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