The equation of line which is parallel to 5x+12y+1 and 5x+12y+7 and lying midway between them
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Answer:
5x−12y−78=0
5x−12y+78=0
Step-by-step explanation:
Given equation of line is
5x−12y+26=0
let the required equation of line parallel to the given line is
5x−12y+c=0
d= 5/2+12/2
∣c−26 = 13∣c−26∣ =4
⇒∣c−26∣=52
⇒c−26=±52
⇒c=−26,78
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