1
MCQ (Single Correct Answer)

JEE Main 2017 (Online) 9th April Morning Slot

A square, of each side 2, lies above the x-axis and has one vertex at the origin. If one of the sides passing through the origin makes an angle 30o with the positive direction of the x-axis, then the sum of the x-coordinates of the vertices of the square is :
A
$$2\sqrt 3 - 1$$
B
$$2\sqrt 3 - 2$$
C
$$\sqrt 3 - 2$$
D
$$\sqrt 3 - 1$$

Explanation



Let, coordinate of point A = (x, y).

$$\therefore\,\,\,$$ For point A,

$${x \over {\cos {{30}^ \circ }}}$$ = $${y \over {\sin {{30}^ \circ }}}$$ = 2

$$ \Rightarrow $$ x = $$\sqrt 3 $$

and y = 1

Similarly, For point B,

$${x \over {\cos {{75}^ \circ }}}$$ = $${y \over {\sin {{75}^ \circ }}}$$ = 2$$\sqrt 2 $$

$$\therefore\,\,\,$$ x = $$\sqrt 3 - 1$$

y = $$\sqrt 3 + 1$$

For point C,

$${x \over {cos{{120}^ \circ }}}$$ = $${y \over {sin{{120}^ \circ }}}$$ = 2

$$ \Rightarrow $$$$\,\,\,$$ x = $$-$$1

y = $$\sqrt 3 $$

$$\therefore\,\,\,$$ Sum of the x - coordinate of the vertices

= 0 + $$\sqrt 3 $$ + $$\sqrt 3 $$ $$-$$ 1 + ($$-$$ 1) = 2$$\sqrt 3 $$ $$-$$ 2
2
MCQ (Single Correct Answer)

JEE Main 2018 (Offline)

A straight line through a fixed point (2, 3) intersects the coordinate axes at distinct points P and Q. If O is the origin and the rectangle OPRQ is completed, then the locus of R is :
A
3x + 2y = 6xy
B
3x + 2y = 6
C
2x + 3y = xy
D
3x + 2y = xy

Explanation



Let coordinate of point R = (h, k).

Equation of line PQ,

(y $$-$$ 3) = m (x $$-$$ 2).

Put y = 0 to get coordinate of point p,

0 $$-$$ 3 = (x $$-$$ 2)

$$ \Rightarrow $$ x = 2 $$-$$ $${3 \over m}$$

$$\therefore\,\,\,$$ p = (2 $$-$$ $${3 \over m}$$, 0)

As p = (h, 0) then

h = 2 $$-$$ $${3 \over m}$$

$$ \Rightarrow $$ $${3 \over m}$$ = 2 $$-$$ h

$$ \Rightarrow $$ m = $${3 \over {2 - h}}$$ . . . . . . (1)

Put x = 0 to get coordinate of point Q,

y $$-$$ 3 $$=$$ m (0 $$-$$ 2)

$$ \Rightarrow $$ y = 3 $$-$$ 2m

$$\therefore\,\,\,$$ point Q = (0, 3 $$-$$ 2m)

And From the graph you can see Q = (0, k).

$$\therefore\,\,\,$$ k = 3 $$-$$ 2m

$$ \Rightarrow $$ m = $${{3 - k} \over 2}$$ . . . . (2)

By comparing (1) and (2) get

$${3 \over {2 - h}} = {{3 - k} \over 2}$$

$$ \Rightarrow $$ (2 $$-$$ h)(3 $$-$$ k) = 6

$$ \Rightarrow $$ 6 $$-$$ 3h $$-$$ 2K + hk = 6

$$ \Rightarrow $$ 3 h + 2K = hk

$$\therefore\,\,\,$$ locus of point R is 3x + 2y= xy
3
MCQ (Single Correct Answer)

JEE Main 2018 (Online) 15th April Morning Slot

In a triangle ABC, coordinates of A are (1, 2) and the equations of the medians through B and C are respectively, x + y = 5 and x = 4. Then area of $$\Delta $$ ABC (in sq. units) is :
A
12
B
4
C
5
D
9

Explanation

Median through C is x = 4

So the coordinate of C is 4. Let C = (4, y), then the midpoint of A(1, 2) and

C(4, y) is D which lies on the median through B.



$$ \therefore $$$$\,\,\,$$ D = $$\left( {{{1 + 4} \over 2},{{2 + y} \over 2}} \right)$$

Now,    $${{1 + 4 + 2 + y} \over 2}$$ = 5  $$ \Rightarrow $$ y = 3.

So, C $$ \equiv $$ (4, 3).

The centroid of the triangle is the intersection of the mesians, Here the medians x = 4 and x + 4 and x + y = 5 intersect at G (4, 1).

The area of triangle $$\Delta $$ABC = 3 $$ \times $$ $$\Delta $$AGC

= 3 $$ \times $$ $${1 \over 2}$$ [1(1 $$-$$ 3) + 4(3 $$-$$ 2) + 4(2 $$-$$ 1)] = 9.
4
MCQ (Single Correct Answer)

JEE Main 2018 (Online) 15th April Evening Slot

The sides of a rhombus ABCD are parallel to the lines, x $$-$$ y + 2 = 0 and 7x $$-$$ y + 3 = 0. If the diagonals of the rhombus intersect P(1, 2) and the vertex A (different from the origin) is on the y-axis, then the coordinate of A is :
A
$${5 \over 2}$$
B
$${7 \over 4}$$
C
2
D
$${7 \over 2}$$

Explanation

Let the coordinate A be (0, c)

Equations of the given lines are

x $$-$$ y + 2 = 0 and 7x $$-$$ y + 3 = 0

We know that the diagonals of the rhombus will be parallel to the angle bisectors of the two given lines; y = x + 2 and y = 7x + 3

$$\therefore\,\,\,$$ equation of angle bisectors is given as :

$${{x - y + 2} \over {\sqrt 2 }} = \pm {{7x - y + 3} \over {5\sqrt 2 }}$$

5x $$-$$ 5y + 10 = $$ \pm $$ (7x $$-$$ y + 3)

$$\therefore\,\,\,$$ Parallel equations of the diagonals are 2x + 4y $$-$$ 7 = 0

and 12x $$-$$ 6y + 13 = 0

$$\therefore\,\,\,$$ slopes of diagonals are $${{ - 1} \over 2}$$ and 2.

Now, slope of the diagonal from A(0, c) and passing through P(1, 2) is (2 $$-$$ c)

$$\therefore\,\,\,$$ 2 $$-$$ c = 2 $$ \Rightarrow $$ c = 0 (not possible)

$$ \therefore $$$$\,\,\,$$ 2 $$-$$ c = $${{ - 1} \over 2}$$ $$ \Rightarrow $$ c = $${5 \over 2}$$

$$\therefore\,\,\,$$ Coordinate of A is $${5 \over 2}$$.

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