InterviewSolution
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InterviewSolution
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Find the distance between the parallel line 15x+8y-34=0 and 15x+8y+31=0. |
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Answer» Converting each of the given equations to the form `y=mx+C`, we get `15x+8y-34=0 Rightarrow y=(-15)/(8)x+(17)/(4).......(i)` `15x+8y+31=0 Rightarrow y=(-15)/(8)x-(31)/(8).......(ii)` Clearly, the slope of the given line are equal and so they are parallel. The given line are of the form `y=mx+C_(1) and y=mx+C_(2)"Where ",m=(-15)/(8), C_(1)=(17)/(4) and C_(2)=(-31)/(8)` `therefore` distance the given lines. `=(|C_(2)-C_(1)|)/(sqrt(1+m^(2))),"where m"=(-15)/(8),C_(1)=(17)/(4) and C_(2)=(-31)/(8)` `=(|(-31)/(8)-(17)/(4)|)/(sqrt(1+((-15)/(8))^(2)))=(|-(65)/(8)|)/(sqrt(1+(225)/(64)))=(((65)/(8)))/(sqrt(1+(289)/(64)))=((65)/(8)xx(8)/(17))=(65)/(17)"units"` Hence, the distance between the given line is `(65)/(17)` units. |
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