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The distance of the point of intersection of the lines `2x-3y+5=0` and `3x+4y=0` from the line `5x-2y=0` isA. `(130)/(17sqrt129)`B. `(13)/(7sqrt29)`C. `(130)/(7)`D. None of these |
Answer» Correct Answer - A Given equation of lines `2x-3y+5=0` and `3x+4y=0` From Eq.(ii),put the volume of `x=(-4y)/(3)` in Eq.(i), we get `2((-4y)/(3))-3y+5=0` `rArr -8y-9y+15=0` `rArry=(15)/(17)` From Eq.(ii), `3x+4.(15)/(17)=0` `rArr x=(-60)/(17.3)=(-20)/(17)` So, the point of intersection is `((-20)/(17),(15)/(17))` `therefore` Required distances form the line `5x-2y=0` is, `d=|-5xx(20)/(17)-2((15)/(17))|/(sqrt(25+4))=|(-100)/(17)-(30)/(17)|/(sqrt29)=(130)/(17sqrt29)` `[because distance of a point `p(x_1y_1)` form the line `ax+by+c=0` is `d =|(ax_1+by_1+c)|/(sqrt(a^2+b^2`]` |
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