Exp.(c)Given(i) A line through (1,2) meets the coordinate axes at P and Q.(ii) The area of △OPQ is minimum.To find The slope of line PQ.Let m be the slope of the line PQ, then the equation of PQ isy−2=m(x−1)Now, PQ meets X-axis at P(1−m2​,0) and Y-axis at Q(0,2−m)⇒OP=1−m2​ and OQ=2−mAlso, area of △OPQ=21​(OP)(OQ)=21​∣∣​(1−m2​)(2−m)∣∣​=21​∣∣​2−m−m4​+2∣∣​=21​∣∣​4−(m+m4​)∣∣​Let f(m)=4−(m+m4​)⇒f′(m)=−1+m24​Now, f′(m)=0⇒m2=4⇒m=±2⇒f(2)=0 and f(−2)=8Since, the area cannot be zero, hence the required value of m is −2.