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Let `d_1` and `d_2` be the lengths of the perpendiculars drawn from any point of the line `7x-9y+10= 0` upon the lines `3x+ 4y =5` and `12x+ 5y= 7` respectively. Then `(A) d_1> d_2 (B) d_1=d_2 (C) d_1< d_2 (D) d_1= 2d_2` |
Answer» `7h-9y+10=0` `9y=7h+10` `y=(7h+10)/9` `d_1=|(3h+4((7h+10)/9)-5)/sqrt(9+16)|` `d_1=|(27h+28h+40-45)/45|` `d_1=|(55h-5)/45|` `d_1=|(11h-1)/9|` `d_2=|(12h+5((7h+10)/9)-7)/sqrt(140+25)|` `d_2=|(143h-13)/(13*9)|` `d_2=|(11h-1)/9|` so, `d_1-d_2`. |
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