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A straight line through A (-15 -10) meet...

A straight line through `A (-15 -10)` meets the lines `x-y-1=0`, `x+2y=5` and `x+3y=7` respectively at A, B and C. If `12/(AB)+40/(AC)=52/(AD)` prove that the line passes through the origin.

Text Solution

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`(x-x_1)/costheta=(y-y_1)/sintheta=r`
`x=rcostheta-15,y=rsintheta-10`
`L_1:r_1costheta-15-r_1sintheta+10-1=0`
`r_1=6/(costheta-sintheta)`
`L_2:r_2costheta-15+2rsintheta-20=5`
`r_2=40/(costheta+2sintheta)`
`L_3:r_3costheta-15+3r_3sintheta-30=7`
`r_3=52/(costheta+3sintheta)`
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