Question

The line $\frac{x}{3}+\frac{y}{4}=1$ meets the $y-$axis and $x-$ axis at $A$ and $B,$ respectively. A square $ABCD$ is constructed on the line segment $AB$ away from the origin. the coordinates of the vertex of the square farthest from the origin are

• A. $(7,3)$
• B. $(4,7)$
• C. $(6,4)$
• D. $(3,8)$

Answer 1

The correct option is B $(4,7)$

The co-ordinates of the vertex of the square farthest from the origin will be the vertex along the diagonal of the square which passes through the origin.

Now consider the equation of the line

$\frac{x}{3}+\frac{y}{4}=1$

Hence it cuts the x axis at $B=(3,0)$ and y axis at $A=(0,4)$

Hence the side of the square AB will be of length 5 units.

Slope of AB is

$=\frac{4-0}{0-3}$

$=\frac{-4}{3}$.

Let the farthest vertex be C.

Then slope of AC will be perpendicular to AB

$=\frac{3}{4}$.

It passes through $(0,4)$

Hence the equation of AC will be

$\frac{y-4}{x}=\frac{3}{4}$

$4y-16=3x$

$y=\frac{3}{4}x+4$

Hence C can be expressed as

$(x,\frac{3x}{4}+4)$

Now the length of AC will be 5 units since the quadrilateral is a square

Hence

$A{C}^2=25$

$(x-0{)}^2+(\frac{3x}{4}+4-4{)}^2=25$

$x^2+\frac{9x^2}{16}=25$

$\frac{25x^2}{16}=25$

$x^2=16$

$x=\pm 4$

Since it is away from the origin, $x=4$

Hence

$y=\frac{3x}{4}+4$

$=3+4$
$=7$

Thus the co-ordinates of Vertex C is

$=(4,7)$.

Thus the co-ordinates of the farthest vertex is $(4,7)$.