InterviewSolution
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InterviewSolution
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Prove that the line 5x-2y-1=0 is midpoint to the line `5x-2y-9=0 and 5x-2y+7=0` |
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Answer» Converting each of the given equations to the form y=mx+C, We get `5x-2y-1=0 Rightarrow y=(5)/(2)x -(1)/(2).......(i)` `5x-2y-9=0 Rightarrow y=(5)/(2)x -(9)/(2).......(ii)` `5x-2y+7=0 Rightarrow y=(5)/(2)x +(7)/(2).......(iii)` Clearly, the slope of (i) is equal to the slope of each of (ii) and (iii), So, the line (i) is parallel to each of the given line (ii) and (iii). Let the given line be `y=mx+C, y=mx+C_(1) and y=mx+C_(2)` respectively. Then, `m=(5)/(2),C=-(1)/(2)=-(1)/(2),C_(1)=-(9)/(2) and C_(2)=(7)/(2)` Let `d_(1) and d_(2)` be the distance of (i) from (ii) and (iii) respectively. `"Then", d_(1)=(|C_(1)-C|)/(sqrt(1+m^(2)))=(-(9)/(2)+(1)/(2))/(sqrt(1+(25)/(4)))=|-4|=(4xx2)/(sqrt29)=(8)/(sqrt29)"units"` `and d_(2)=(|C_(2)-C|)/(sqrt(1+m^(2)))=((7)/(2)+(1)/(2))/(sqrt(1+(25)/(4)))=(4xx(2)/(sqrt29))=(8)/(sqrt29)"units"` Thus, `d_(1)=d_(2)` This shows that (i) is equidistant from (ii) and (iii). Hence, 5x-2y-1=0 is mid-parallel to the lines 5x-2y-9=0 and 5x-2y+7=0 |
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