A=(3,1) Let slope of line be m ∴‌‌y−y1=m(x−x1)‌‌ [be the required line] y−1=m(x−3) y−1=mx−3m mx−y+(1−3m)=0....(i)
The greatest distance of line from origin passes through A(3,1) is perpendicular to the given line. ∴‌‌OA⟂PQ Slope of OA× Slope of PQ=−1 ‌
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×m=−1⇒m=−3 Put, m=−3 in Eq. (i) −3x−y+(1−3)(−3)=0 −3x−y+(10)=0 3x+y+10=0 ∴ Area of