Question

One diagonal of a square is the portion of the straight line $3x+4y=12$ intercepted between the coordinate axes. If the extremities of the other diagonal are $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$, then find the value of $\left(x_1^2+x_2^2+y_1^2+y_2^2\right)$.

Answer 1

Answer: 0025

Let the diagonal $AC$ of the square $ABCD$ be along the line $3x+4y=12$
$\therefore A=(4,0)$ and $C=(0,3)$
The coordinates of midpoint $M$ of $AC$ are $\left(2,\frac{3}{2}\right)$.
$AsBD\perp AC$
$\therefore$ Slope of $BD=\frac{4}{3}=\tan \theta$, where $\theta$ is the angle which $BD$ makes with the positive direction of $x$-axis.
$\therefore$ The equation of BD in parametric form is
$\frac{x-2}{3/5}=\frac{y-\frac{3}{2}}{4/5}=r(\text{ let })\left(\text{ As }\tan \theta =\frac{4}{3}\right)$
Put $r=\pm \frac{5}{2}$ in equation (1), we get
$\therefore x=2+\frac{3}{5}\left(\pm \frac{5}{2}\right)$ and $y=\frac{3}{2}+\frac{4}{5}\left(\pm \frac{5}{2}\right)$
$\therefore$ Co-ordinates of other diagonal are $\left(\frac{7}{2},\frac{7}{2}\right)$ and $\left(\frac{1}{2},\frac{-1}{2}\right)$
Hence the value of expression $x_1^2+x_2^2+y_1^2+y_2^2=\frac{49}{4}+\frac{1}{4}+\frac{49}{4}+\frac{1}{4}=\frac{100}{4}=25$