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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 triangle OPQ is least, then the slope of the line PQ is :

Option 1)
$-\frac{1}{4}$

Option 2)
$- 4$

Option 3)
$-2$

Option 4)
$-\frac{1}{2}$

Answers (2)

best_answer

As learnt in

Slope – intercept from of a straight line -

$y=mx+C$

- wherein

$m\rightarrow$ slope

$c\rightarrow y-$ intercept of the line .

Apply EQUATION OF A STRAIGHT LINE IN Slope -intercept form -

This line passes through the point (1,2) so

$y=mx+c \Rightarrow 2=m+c$

$\Rightarrow c=2-m$

Now, x co-ordinate of this line would be $\dfrac{m-2}{m}$

and y co-ordinate of this line would be $2-m$

Hence Area

$A=\dfrac{1}{2}* \left | (2-m)\right |*\left | \dfrac{(m-2)}{m} \right |$

$\Rightarrow A=\dfrac{(m-2)^2}{2m}$

We have to find the value of m for which A is least. So,

$\Rightarrow \frac{m^{2}-4m+4}{2m}= A\Rightarrow \frac{m}{2}-2+\frac{2}{m}= A$

$\therefore \frac{dA}{dm}= 0$

$\Rightarrow \frac{1}{2}-\frac{2}{m^{2}}= 0\Rightarrow \frac{1}{2}= \frac{2}{m^{2}}\Rightarrow m^{2}= 4\Rightarrow m= \pm 2$

Option 1)

$-\frac{1}{4}$

This option is incorrect.

Option 2)

$- 4$

This option is incorrect.

Option 3)

$-2$

This option is correct.

Option 4)

$-\frac{1}{2}$

This option is incorrect.

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