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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