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Let the lines 2x+3y+19=0 and 9x+6y-17=0 cut the x-axis in A, B and y-axis in C, D then prove that OA.OB=OC.OD (where Ois the origin) and hence prove A,B,C,D are concyclic.

Let the lines 2x+3y+19=0 and 9x+6y-17=0 cut the x-axis in A, B and y-axis in C, D then prove that OA.OB=OC.OD (where Ois the origin) and hence prove A,B,C,D are concyclic.

Grade:12

1 Answers

Askiitians Expert Bharath-IITD
23 Points
14 years ago

Dear Harshit,

for the equation 2x+3y+19=0

the point of intersection of this line with x axis is  founby equating y=0

so 2x = -19   --> x = -19/2

So the point A is ( -19/2 , 0)

similarily we find B as ( 17/9 , 0 ) for the other equation ,

and also C is ( 0, -19/3 ) and D is (0, 17/6 )

So we have OA= 19/2, OB = 17/9, OC = 19/3 and OD = 17/6

so OA * OB =  (19 * 17 )/18

And OC * OD = ( 19*17 )/18

so first condition is proved i.e, OA * OB = OC * OD

Now

if the points are concyclic the the perpendicular bisectors of all the the three chords formed by these lines as shown in the figure should meet at one point  

 

So the euation of the line through

we find the equations of the perpendicular bisectors of the lines AD , DB, BC and check whether all the three lines have the same common point of intersection. It comes out to be common  and can be proved. Also in this case we can find the determinant of the matrix formed by the coefficients of the perpendicular bisector lines then if its is zero then they are concurrent i.e. at the centre of the circle on which they lie.

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Regards,

Askiitians Experts

Adapa Bharath

 

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