Question

Statement-1: Area of triangle formed by the line which is passing through the point \[\left( 5,6 \right)\] such that segment of the line between the axes is bisected at the point; with coordinate axes is \[60\] sq. units.
Statement-2: Area of the triangle formed by the line passing through point \[\left( \alpha ,\beta \right)\] with axes is minimum when point \[\left( \alpha ,\beta \right)\] is the midpoint of segment of line between axes.
A) Statement-1 is true; Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
B) Statement-1 is true; Statement-2 is true; Statement-2 is NOT a correct explanation for Statement-1.
C) Statement-1 is true; Statement-2 is false.
D) Statement-1 is false; Statement-2 is true.

Answer 1

Hint: Check the validity of the second statement by writing the equation of line using one point form and finding the points of intersection of the line with coordinates axes. Find the area of the triangle and minimize it by differentiating it to zero. Solve the equations to find the relation between slope of line and the given point.

Complete step-by-step answer:
We will begin by checking the validity of statement-2. We have a point \[\left( \alpha ,\beta \right)\]. We have to draw a line passing through this point.
Let’s assume that the slope of this line is \[m\].
Using one point form of line, we know that the equation of any line passing through \[\left( \alpha ,\beta \right)\] and having slope \[m\] is \[y-\beta =m(x-\alpha )\].
We will now find the point of intersection of this line with coordinates axes.
To find a point on the \[x-\] axis, substitute \[y=0\] in the equation of line.
Thus, we have \[0-\beta =m(x-\alpha )\].
Rearranging the terms, we get \[x=\dfrac{-\beta }{m}+\alpha \].
Let this point be \[A\left( \dfrac{-\beta }{m}+\alpha ,0 \right)\] on the \[x-\] axis.
To find a point on the \[y-\] axis, substitute \[x=0\] in the equation of the line.
Thus, we have \[y-\beta =m(0-\alpha )\].
Rearranging the terms, we get \[y=-m\alpha +\beta \].
Let this point be \[B\left( 0,\beta -m\alpha \right)\] on the \[x-\] axis.
We know that the area of triangle formed by a line intersecting the coordinates axes at points \[\left( a,0 \right)\] and \[\left( 0,b \right)\] is \[\dfrac{1}{2}ab\].
Substituting \[a=\dfrac{-\beta }{m}+\alpha ,b=\beta -m\alpha \] in the above equation, the area of triangle is \[\dfrac{1}{2}\left( \dfrac{-\beta }{m}+\alpha \right)\left( \beta -m\alpha \right)=\dfrac{1}{2}\left( \dfrac{-{{\beta }^{2}}}{m}+2\beta \alpha -m{{\alpha }^{2}} \right)\].
We have to find the minimum area of the triangle. Thus, we will differentiate with respect to slope \[m\] and equate it to zero.
Thus, we have \[\dfrac{d}{dm}\left( \dfrac{1}{2}\left( \dfrac{-{{\beta }^{2}}}{m}+2\beta \alpha -m{{\alpha }^{2}} \right) \right)=-\dfrac{{{\alpha }^{2}}}{2}+\dfrac{{{\beta }^{2}}}{2{{m}^{2}}}\] using the fact that if \[y=a{{x}^{n}}\] then \[\dfrac{dy}{dx}=na{{x}^{n-1}}\] and that differentiation of a constant is zero.
Now, we have \[\dfrac{d}{dm}\left( \dfrac{1}{2}\left( \dfrac{-{{\beta }^{2}}}{m}+2\beta \alpha -m{{\alpha }^{2}} \right) \right)=-\dfrac{{{\alpha }^{2}}}{2}+\dfrac{{{\beta }^{2}}}{2{{m}^{2}}}=0\].
\[\begin{align}
  & \Rightarrow -\dfrac{{{\alpha }^{2}}}{2}+\dfrac{{{\beta }^{2}}}{2{{m}^{2}}}=0 \\
 & \Rightarrow m=\pm \dfrac{\beta }{\alpha } \\
\end{align}\]
We will check the value of \[m\] that will give the minimum value. Thus, to get the minimum value, differentiate the area of the triangle twice and substitute the value of \[m\] to check if the value is greater than or less than zero. If the value we get after substituting the value of \[m\] in the double differentiation of the area, then that value of \[m\] corresponds to the minimum area.
Thus, we have \[\dfrac{{{d}^{2}}}{d{{m}^{2}}}\left( \dfrac{1}{2}\left( \dfrac{-{{\beta }^{2}}}{m}+2\beta \alpha -m{{\alpha }^{2}} \right) \right)=\dfrac{d}{dm}\left( -\dfrac{{{\alpha }^{2}}}{2}+\dfrac{{{\beta }^{2}}}{2{{m}^{2}}} \right)=\dfrac{-{{\beta }^{2}}}{{{m}^{3}}}\].
The value \[\dfrac{-{{\beta }^{2}}}{{{m}^{3}}}\] is positive for negative values of \[m\].
Thus, \[m=\dfrac{-\beta }{\alpha }\] corresponds to the slope of the line which gives the minimum area of the triangle.
We will now find the midpoint of the points \[A\left( \dfrac{-\beta }{m}+\alpha ,0 \right)\] and \[B\left( 0,\beta -m\alpha \right)\].
Substituting the value \[m=\dfrac{-\beta }{\alpha }\] in the coordinates of point \[A\left( \dfrac{-\beta }{m}+\alpha ,0 \right)\], we get \[A\left( \dfrac{-\beta }{m}+\alpha ,0 \right)=\left( 2\alpha ,0 \right)\].
Substituting the value \[m=\dfrac{-\beta }{\alpha }\] in the coordinates of point \[B\left( 0,\beta -m\alpha \right)\], we get \[B\left( 0,\beta -m\alpha \right)=\left( 0,2\beta \right)\].
We will now find the midpoint of points \[A\left( 2\alpha ,0 \right)\] and \[B\left( 0,2\beta \right)\].
We know that midpoint of two points with coordinates \[\left( x,y \right)\] and \[\left( u,v \right)\]is\[\left( \dfrac{x+u}{2},\dfrac{y+v}{2} \right)\].
Substituting \[x=2\alpha ,y=0,u=0,v=2\beta \] in the above equation, we have \[\left( \dfrac{2\alpha +0}{2},\dfrac{0+2\beta }{2} \right)=\left( \alpha ,\beta \right)\].
Hence, we have Statement-2 as true.
Now, we will check Statement-1.
Substituting \[\alpha =5,\beta =6\] in above situation, we have area of triangle as \[\dfrac{1}{2}\left( 2\alpha \right)\left( 2\beta \right)=2\alpha \beta =2\times 5\times 6=60\]sq. units.
Hence, Statement-1 and Statement-2 are true and Statement-2 is the correct explanation of Statement-1.
Note: We can’t comment anything about Statement-1 without checking the validity of Statement-2. Hence, it’s necessary to first check if Statement-2 is correct or not.