distance of the lines `2x-3y-4=0` from the point `(1, 1)` measured paralel to the line `x+y=1` isA. `sqrt2`B. `(5)/(sqrt2)`C. `(1)/(sqrt2)`D. 6

distance of the lines `2x-3y-4=0` from the point `(1, 1)` measured par

1.	distance of the lines `2x-3y-4=0` from the point `(1, 1)` measured paralel to the line `x+y=1` isA. `sqrt2`B. `(5)/(sqrt2)`C. `(1)/(sqrt2)`D. 6
Answer» The equation of the line through the point P(1,1) parallel to the line x + y = 1 is `(x -1)/(cos (3pi)/(4)) = (y-1)/(sin (3pi)/(4))` Suppose it meets the line 2 x - 3y = 4 at `Q (1 + r cos (3pi)/(4) , 1 + r sin (3pi)/(4))` , where PQ = \|r\| As Q lies on the line 2x - 3y = 4 . `therefore 2 (1 - (r)/(sqrt2)) - 3 (1 + (r)/(sqrt2) ) = 4` `implies -1- (5r)/(sqrt2) =4 implies (5r)/(sqrt2) = -5 implies r = - sqrt2` Hence , PQ = `sqrt2`

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distance of the lines `2x-3y-4=0` from the point `(1, 1)` measured paralel to the line `x+y=1` isA. `sqrt2`B. `(5)/(sqrt2)`C. `(1)/(sqrt2)`D. 6

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