IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

chimanshu_007

any line through (2,3) is given by

x - 2 / cos@ = y - 3 / sin@ = r

let this line cuts 2x + y = 5 and 2x + y = 3 at D and C resp. (look in the figure)

we are given DC = 2

D = 2 + r cos@ , 3 + r sin@

C = ( 2 + (r + 2)cos@ , 3 + (r + 2)sin@)

As D , C lie on 2x + y = 5 , 2x + y = 3

2(2 + rcos@) + (3 + rsin@) = 5

2 ( 2 + (r + 2) cos@) + (3 + (r + 2) sin@) = 3

Subtract the above 2 =ns

4 cos@ + 2 sin@ =- 2

2cos@ + sin@ = -1

2 [( 1 - tan²(@/2)/(1 + tan²(@/2)] + [2tan (@/2) / (1 + tan²(@/2))] = -1

Solve this equation , u will get

Tan (@/2) = -1 or tan(@/2) = 3

So tan@ = infinity or -3/4

So the lines

Y - 3 = infinity ( x - 2) and y - 3 = (-3/4)(x-2)

X - 2 = 0 or 3x + 4y - 18 = this are the required equations