Given:
A line through (1,2) meets the coordinate axes at P and O. Also, the area of ΔOPQ is minimum
To find:
The slope of the line PQ
Solution:
Let m be the line PQ, then the equation of PQ is:
y-2=m(x-1)
PQ meets x-axis at P(1-2/m,0) and y-axis at Q(0,2-m)
OP = 1-2/m
OQ = 2-m
Area of ΔOPQ = 1/2(OP)(OQ)
=1/2|(1-2/m)(2-m)|
=1/2|4-(m+4/m)|
Let f(m) = 4-{m+4/m)
f'(m) = -1+4/m<sup>2</sup>
f'(m)=0
m<sup>2</sip>=4
m=±2
f(2)=0 and f(-2)=8
We know that the area cannot be zero, therefore, the required value of m is -2.