Minimum area of a triangle Question: A line is drawn through the point (1, 2) to meet the coordinate axis at P And Q such that OPQ is a triangle with O as the origin. If area of triangle OPQ is least, then what is the slope of line PQ.
I have been given four options
A(-1/4), B(-4), C(-2), D (-1/2)
So I found all four equations of line with given slopes, found the length of intercepts of each and then found the area of each triangle (b = base, h = height) using
$$area(b, h) = \frac1{2}bh$$
My answer came out to be slope = -2
But I was wondering if this is least area out of all possible slopes  of line passing throught (1, 2) or just these four slopes, also is there any concept which we can use to directly obtain the slope of least area without checking each option.
Additional information: I have found slope 1.1 which gives area less than given by slope -2.
 A: Your answer is correct, notice
Let the line intersect the coordinate axes at $P(s, 0)$ & $Q(0, t)$ respectively. where $s$ & $t$ are variables. 
The equation of the line is given as $$\frac{x}{s}+\frac{y}{t}=1$$
Since, the line PQ also passes through the point $(1, 2)$, hence satisfying the equation of the line we get $$\frac{1}{s}+\frac{2}{t}=1$$ $$t=\frac{2s}{s-1}\tag 1$$
Since, $\angle POQ=90^\circ$ hence, the area of the right triangle $POQ$ $$A=\frac{1}{2}(\text{base})(\text{normal height})=\frac{1}{2}(s)(t)$$
setting the value of $t$ from (1), we get $$A=\frac{1}{2}s\left(\frac{2s}{s-1}\right)=\frac{s^2}{s-1}$$ Now, differentiating A w.r.t. $s$, we get $$\frac{dA}{ds}=\frac{(s-1)(2s)-s^2(1)}{(s-1)^2}=\frac{s^2-2s}{(s-1)^2}$$
$$\frac{d^2A}{ds^2}=\frac{(s-1)^2(2s-2)-2(s^2-2s)(s-1)}{(s-1)^4}=\frac{2}{(s-1)^3}$$ For minima & maxima $$\frac{dA}{ds}=0\implies \frac{s^2-2s}{(s-1)^2}=0 $$ $$s(s-2)=0\implies s=0, 2$$ setting $s=2$ in $\frac{d^2A}{ds^2}$, we get $$\frac{d^2A}{ds^2}=\frac{2}{(2-1)^3}=2>0$$ The area A is minimum at $s=2$, now setting $s=2$ in (1), we get $$t=\frac{2(2)}{2-1}=4$$ Now, the equation of the line PQ is given as $$\frac{x}{2}+\frac{y}{4}=1$$ $$\color{blue}{y=-2x+4}$$
Now, comparing the above equation with $y=mx+c$, the slope (m) of the line $PQ$ is $\color{red}{-2}$
A: Like Gabor,
$$\dfrac{y-2}{x-1}=m\iff\dfrac x{(2-m)/m}+\dfrac y{2-m}=1$$
$$\implies A=\left|\dfrac{(2-m)^2}{-2m}\right|$$
Now if $m>0,$ $$\dfrac{(2-m)^2}m=\dfrac4m+m-4\ge2\sqrt{\dfrac4m\cdot m}-4\ge0$$ the equality occurs of $m=2$
if $m<0, n=-m$(say),
$$\dfrac{(2-m)^2}{-m}=\dfrac{(2+n)^2}n=\dfrac4n+n+4\ge2\sqrt{\dfrac4n\cdot n}+4\ge8$$ the equality occurs of $n=2$
A: This is essentially the same problem as (1) $ y = mx +b$ subject to (2) $2 = m+b$, where $m$ is the slope coefficient and $b$ is the y intercept. From (2)  $b =  2-m$.
Rewriting (1): $y = m(x-1) +2$
Now you need to find the x and y intercepts:
x intercept can be found by taking y = 0: 
$ 0 = m(x-1)+2$ $\Longrightarrow$ $ x_0 = 1-\frac{2}{m}$
y intercept can be found by setting x = 0:
$ y = m(-1) +2 $ $\Longrightarrow$ $ y_0 = 2-m$
The area of the triangle will be the following:
A = $\frac{x_0*y_0}{2}$
We need to find $Min (A)$
Rewriting A in terms of m gives:
$\frac{(1-\frac{2}{m})*(2-m)}{2}$
$\frac{d}{dm}$$A$ = $\frac{2}{m^2} - \frac{1}{2}$ $\Longrightarrow$ m = +/- 2 are the local maximum and minimum. The second derivative for m = -2 is positive, hence it is the local minimum.
Min(A) = $\frac{(1-\frac{2}{m})*(2-m)}{2} = 4$ $\Longrightarrow$ m = -2
A: We can determine the triangle completely with only an x-intercept, since the second condition that the line must go through (1,2) will give a complete line.
Note that $x>1$ otherwise we will not get a triangle between the coordinate axes. 
We need an expression for the height of the triangle given $x$, which is the base. If $x>1$ the y-intercept is found by finding the gradient: $$m=\frac{2}{1-x}$$
Then using $y = mx+b$ and $(x,y)=(1,2)\implies$
$$ b = 2+\frac{2}{x-1}$$
Where $b$ is the y-intercept, which is also the height of the triangle.
So the area of the triangle will be $$A = \frac{xb}{2}=\frac{2x+\frac{2x}{x-1}}{2}=x+\frac{x}{x-1}=x+\frac{x-1+1}{x-1}=x+1+\frac{1}{x-1}$$
To find the minimum we differentiate and find where $\frac{dA}{dx}=0$, then we determine whether or not this is a minimum either by finding the second derivative or substituting points. We will assume that it is a minimum in this case.
$$\frac{dA}{dx}=1-\frac{1}{(x-1)^2}$$
Note that this $\frac{dA}{dx}=0$ when $x=2$ (ignoring $x = 0$ because it is not in our domain), and we substitute that into our first formula:
$$m=\frac{2}{1-2}=-2$$
As required.
