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Question

A line is drawn through the point (1,2) to meet the coordinate axes at P and Q such that it forms a triangle OPQ, where O is the origin, if the area of the OPQ is least, then the slope of the line PQ is?


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Solution

Step 1: Find the equation of line PQ in intercept form.

Draw a figure based on given data as shown.

Let m be the slope of the line PQ

Then, the equation of PQ using the given point (1,2) and slope m is

y-2=m(x-1)

⇒y-2=mx-m⇒mx-y=m-2⇒xm-2m-ym-2=1

Step 2: Find the Area of ΔOPQ

From the figure,

PQ meets x-axis at Pm-2m,0 and y-axis at Q0,2-m

x-intercept=OP=m-2m

y-intercept=OQ=2-m

Area of ΔOPQ=12OP×OQ

=12m-2m×2-m=121-2m×2-m=122-m-4m+2=124-m+4m

Step 3: Find the value of m

Let fm=4-m+4m

Differentiate with respect to m.

⇒f'm=-1-4m2

The area of the OPQ is least.

Therefore f'(m)=0

⇒-1+4m2=0⇒m2=4⇒m=2orm=-2

If m=2the area is 0 which is not possible.

Therefore, the required slope of PQ is -2.


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