Question

One diagonal of a square is the portion of $\frac{x}{97}+\frac{y}{79}=1$ intercepted between the axes. $p_1,p_2$ are the lengths of the perpendiculars from the vertices of the other diagonal on the axis of $y$ then $p_1^2+p_2^2$ is equal to

Answer 1

Let $AB$ be the line whose equation is
$\frac{x}{97}+\frac{y}{79}=1$ where $A$ is $(97,0)$ and $B$ is
$(0,79)$. Equation of the other diagonal passing through
$(\frac{97}{2},\frac{79}{2})$, the
mid-point of $AB$ and perpendicular to $AB$ is
$y-\frac{79}{2}=\frac{97}{79}(x-\frac{97}{2})$ ( 1 )
Let
the coordinates of the vertex $C$ of the other diagonal be
$(x_1,y_1)$ then $(x_1,y_1)$ lies on ( 1 ) and $B{C}^2=A{B}^2/2$
$\Rightarrow 2[x_1^2+(y_1-79{)}^2]=(97{)}^2+(79{)}^2$
$\Rightarrow 2[x_1^2+{\{\frac{97}{79}(x_1-\frac{97}{2})-\frac{79}{2}\}}^2]=(97{)}^2+(79{)}^2$
$\Rightarrow 2[(1+\frac{{97}^2}{{79}^2})x_1^2-2\times \frac{97}{79}\left(\frac{{97}^2+{79}^2}{2\times 79}\right)x_1+{\left(\frac{(97{)}^2+(79{)}^2}{2\left(79\right)}\right)}^2]$
$=(97{)}^2+(79{)}^2$
$\Rightarrow 4x_1^2-4\times 97x_1+(97{)}^2+(79{)}^2=2\times (79{)}^2$
$\Rightarrow {(2x_1-97)}^2=(79{)}^2$
$\Rightarrow 2x_1=97\pm 79\Rightarrow x_1=\frac{97\pm 79}{2}{y}_1=\frac{97\pm 79}{2}$
$\therefore$ The coordinates of the vertices of the other diagonal are $(88,88)$ and $(9,9)$
$\Rightarrow p_1=88,p_2=9$
$\therefore p_1^2+p_2^2=7444+81=7825$