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If `(a,b)` and `(c,d)` are two points on the whose equation is `y=mx+k`, then the distance between `(a,b)` and `(c,d)` in terms of `a,c` and `m` isA. `|a+c|sqrt(1+m^(2))`B. `|a-c|sqrt(1+m^(2))`C. `(|a-c|)/(sqrt(1+m^(2))`D. `|a-c|(a+m^(2))` |
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Answer» Correct Answer - B Let `m=tantheta` then equation of line through `(a,b)` is `(x-a)/(costheta)=(y-b)/(sintheta)=r` if `(c,d)` lies on it at `r` distance then `impliesc-a=rcosthetaimpliesr-(c-a)sectheta` `impliesr=|c-a|sqrt(1+m^(2))` Alter: Since `(a,b)` and `(c,d)` are on the same line `y=mx+k`, they satisfy the same equation. Therefore `b=ma+kd=mc+k`. Now the distance between `(a,b)` and `(c,d)` is `sqrt((a-c)^(2)+(b-d)^(2))` From the first to equations we obtain `(b-d)=m(a-c)`, so that `sqrt((a-c)^(2)+(b-d)^(2))=sqrt((a-c)^(2)+m^(2)(a-c)^(2))` `=|a-c|sqrt(1+m^(2))` Note we are using the fact that `=sqrt(x^(2))=|x|` for all real `x`. |
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