Area of triangle(Co-ordinate Geometry) Here's the question: A straight line passing through P(3,1) meet the co-ordinate axes at 'A' & 'B'. It is given that distance of this straight line is maximum from origin. Area of ∆OAB is equal to? Here's what I have done: since it passes through P(3,1) the equation of the line must be x+3y-3=0, then calculated the maximum distance from origin which turned out to be 3/√10, then the points at which it cuts the co-ordinate axes will be (0,1)&(3,0) and distance between them will be √10, but I am not getting the answer.

Let the line be: $\frac{y-1}{x-3} = m$

or, $mx -y + (1-3m) = 0$

Distance of this line from the origin is : $d_0 = \frac{|1-3m|}{\sqrt{m^2+1}}$

![enter image description here](

The function attains _global maxima_ at $m = -3$.

As $d_0$ attains maxima for $m = -3$, the equation of the line now becomes: $3x+y-10=0$. This line intersects the co-ordinate axes at $(\frac{10}{3},0)$ and $(0,10)$, which are the values of $A$ and $B$ (in no particular order).

It follows that the area of $∆OAB$ is $\frac{50}{3}$.

**NOTE** : The function attains _global maxima_ at $m = -3$, i.e., the maximum possible value for $d_0$ is $\sqrt{10}$. The part of the above plot beyond $m = \frac{1}{3}$ belongs to the function $\frac{3m-1}{\sqrt{m^2+1}}$. Although this function is monotonically increasing for $m > \frac{1}{3}$, it never crosses $\sqrt{10}$.

As, $\lim_{x\to\infty} \frac{3m-1}{\sqrt{m^2+1}} = 3 < \sqrt{10}$.