JEE Main 2019 — Mathematics Coordinate Geometry

Given, the two adjacent vertices of the rectangle are $(-8,5)$(−8,5) and $(6,5),$(6,5), the $y$y-co-ordinates of both of these points is same this means they lie on the line $y=5,$y=5, which is parallel to the $x$x-axis.

And, we know that the sides of a rectangle are perpendicular to each other, hence the other two sides passing through these vertices are parallel to $y$y-axis.

Also, we know that the equation of a line parallel $y$y-axis is $x=k.$x=k.

Therefore, the equation of the sides of the rectangle passing through $(-8,5)$(−8,5) and $(6,5)$(6,5) are respectively $x=-8$x=−8 and $x=6.$x=6.

And, hence the other two vertices of the rectangle can be taken as $(6,a)$(6,a) and $(-8,b).$(−8,b).

And since the other side is parallel to $x$x-axis, hence $a=b.$a=b.

Here, diagonal subtends right angle at the circumference of circle, so its mid-point is the same as centre of circle, which lies on the given diameter.

The mid-point of a line segment joining the points $({x}_{1},{y}_{1})$(x1​,y1​) and $({x}_{2},{y}_{2})$(x2​,y2​) is $(\frac{{x}_{1}+{x}_{2}}{2},\frac{{y}_{1}+{y}_{2}}{2}),$(2x1​+x2​​,2y1​+y2​​), hence, the mid-point of the diagonal of the rectangle is $(\frac{-8+6}{2},\frac{a+5}{2})=(-1,\frac{a+5}{2})$(2−8+6​,2a+5​)=(−1,2a+5​)

This point lie on the diameter $3y=x+7$3y=x+7

So, $\frac{3(a+5)}{2}=-1+7$23(a+5)​=−1+7

$$\Rightarrow a+5=\frac{2(6)}{3}$$⇒a+5=32(6)​

$$\Rightarrow a=-1$$⇒a=−1

Hence, the vertices are $(-8,5),(6,5),(6,-1)$(−8,5),(6,5),(6,−1) and $(-8,-1).$(−8,−1).

The distance between the points $({x}_{1},{y}_{1})$(x1​,y1​) and $({x}_{2},{y}_{2})$(x2​,y2​) is $\sqrt{{({x}_{1}-{x}_{2})}^{2}+{({y}_{1}-{y}_{2})}^{2}}$(x1​−x2​)2+(y1​−y2​)2​

Thus, the length of sides are $6$6 and $14$14 units.

Hence, area $=l\times b=14\times 6=84$=l×b=14×6=84 sq. units