Joint Entrance Examination

Graduate Aptitude Test in Engineering

Geomatics Engineering Or Surveying

Engineering Mechanics

Hydrology

Transportation Engineering

Strength of Materials Or Solid Mechanics

Reinforced Cement Concrete

Steel Structures

Irrigation

Environmental Engineering

Engineering Mathematics

Structural Analysis

Geotechnical Engineering

Fluid Mechanics and Hydraulic Machines

General Aptitude

1

A straight line through origin O meets the lines 3y = 10 − 4x and 8x + 6y + 5 = 0 at points A and B respectively. Then O divides the segment AB in the ratio :

A

2 : 3

B

1 : 2

C

4 : 1

D

3 : 4

The lines 4x + 3y $$-$$ 10 = 0 and

8x + 6y + 5 = 0 , are parallel as

$${4 \over 8}$$ = $${3 \over 6}$$

Now length of perpendicular from

(0, 0, 0) to 4x + 3y $$-$$ 10 = 0 is,

P_{1} = $$\left| {{{4\left( 0 \right) + 3\left( 0 \right) - 10} \over {\sqrt {{4^2} + {3^2}} }}} \right|$$ = $${{10} \over 5}$$ = 2

Length of perpendicular from

0 (0, 0) to 8x + 6y + 5 = 0 is

P_{2} = $$\left| {{{8\left( 0 \right) + 6\left( 0 \right) + 5} \over {\sqrt {{6^2} + {8^2}} }}} \right|$$ = $${5 \over {10}}$$ = $${1 \over 2}$$

$$\therefore\,\,\,$$ P_{1} : P_{2} = 2 : $${1 \over 2}$$ = 4 : 1

8x + 6y + 5 = 0 , are parallel as

$${4 \over 8}$$ = $${3 \over 6}$$

Now length of perpendicular from

(0, 0, 0) to 4x + 3y $$-$$ 10 = 0 is,

P

Length of perpendicular from

0 (0, 0) to 8x + 6y + 5 = 0 is

P

$$\therefore\,\,\,$$ P

2

Let k be an integer such that the triangle with vertices (k, – 3k), (5, k) and (–k, 2) has area 28 sq. units. Then the orthocentre of this triangle is at the point:

A

$$\left( {1,{3 \over 4}} \right)$$

B

$$\left( {1, - {3 \over 4}} \right)$$

C

$$\left( {2,{1 \over 2}} \right)$$

D

$$\left( {2, - {1 \over 2}} \right)$$

Given, vertices of triangle are (k, – 3k), (5, k) and (–k, 2).

$${1 \over 2}\left| {\matrix{ k & { - 3k} & 1 \cr 5 & k & 1 \cr { - k} & 2 & 1 \cr } } \right| = \pm 28$$

$$ \Rightarrow $$ k(k - 2) + 3k(5 + k) + 1(10 + k^{2}) = $$ \pm $$ 56

$$ \Rightarrow $$ 5k^{2} + 13k + 10 = $$ \pm $$ 56

$$ \Rightarrow $$ 5k^{2} + 13k - 66 = 0

$$ \Rightarrow $$ k = $${{ - 13 \pm \sqrt { - 1151} } \over {10}}$$

So no real solution exist.

or 5k^{2} + 13k - 46 = 0

$$ \therefore $$ k = $${{ - 23} \over 5}$$ or k = 2

since k is an integer $$ \therefore $$ k = 2

Thus, the coordinate of vertices of triangle are

A(2, -6), B(5, 2) and C(-2, 2).

Now, equation of altitude from vertex A is

y - (-6) = $${{ - 1} \over {\left( {{{2 - 2} \over { - 2 - 5}}} \right)}}\left( {x - 2} \right)$$

$$ \Rightarrow $$ x = 2 .......(1)

Equation of altitude from vertex B is

y - 2 = $${{ - 1} \over {\left( {{{2 + 6} \over { - 2 - 2}}} \right)}}\left( {x - 5} \right)$$

$$ \Rightarrow $$ 2y - 4 = x - 5

$$ \Rightarrow $$ x - 2y = 1 .......(2)

Point H($$\alpha $$, $$\beta $$) lies on both (1) and (2),

$$ \therefore $$ $$\alpha $$ = 2 .........(3)

$$\alpha $$ - 2$$\beta $$ = 1 ......(4)

Solving (3) and (4), we get

$$\alpha $$ = 2 , $$\beta $$ = $${1 \over 2}$$

$$ \therefore $$ Orthocentre is $$\left( {2,{1 \over 2}} \right)$$.

$${1 \over 2}\left| {\matrix{ k & { - 3k} & 1 \cr 5 & k & 1 \cr { - k} & 2 & 1 \cr } } \right| = \pm 28$$

$$ \Rightarrow $$ k(k - 2) + 3k(5 + k) + 1(10 + k

$$ \Rightarrow $$ 5k

$$ \Rightarrow $$ 5k

$$ \Rightarrow $$ k = $${{ - 13 \pm \sqrt { - 1151} } \over {10}}$$

So no real solution exist.

or 5k

$$ \therefore $$ k = $${{ - 23} \over 5}$$ or k = 2

since k is an integer $$ \therefore $$ k = 2

Thus, the coordinate of vertices of triangle are

A(2, -6), B(5, 2) and C(-2, 2).

Now, equation of altitude from vertex A is

y - (-6) = $${{ - 1} \over {\left( {{{2 - 2} \over { - 2 - 5}}} \right)}}\left( {x - 2} \right)$$

$$ \Rightarrow $$ x = 2 .......(1)

Equation of altitude from vertex B is

y - 2 = $${{ - 1} \over {\left( {{{2 + 6} \over { - 2 - 2}}} \right)}}\left( {x - 5} \right)$$

$$ \Rightarrow $$ 2y - 4 = x - 5

$$ \Rightarrow $$ x - 2y = 1 .......(2)

Point H($$\alpha $$, $$\beta $$) lies on both (1) and (2),

$$ \therefore $$ $$\alpha $$ = 2 .........(3)

$$\alpha $$ - 2$$\beta $$ = 1 ......(4)

Solving (3) and (4), we get

$$\alpha $$ = 2 , $$\beta $$ = $${1 \over 2}$$

$$ \therefore $$ Orthocentre is $$\left( {2,{1 \over 2}} \right)$$.

3

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 30^{o} 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

$$\sqrt 3 - 1$$

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

4

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

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

Number in Brackets after Paper Name Indicates No of Questions

AIEEE 2002 (4) *keyboard_arrow_right*

AIEEE 2003 (5) *keyboard_arrow_right*

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JEE Main 2013 (Offline) (2) *keyboard_arrow_right*

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Trigonometric Functions & Equations *keyboard_arrow_right*

Properties of Triangle *keyboard_arrow_right*

Inverse Trigonometric Functions *keyboard_arrow_right*

Complex Numbers *keyboard_arrow_right*

Quadratic Equation and Inequalities *keyboard_arrow_right*

Permutations and Combinations *keyboard_arrow_right*

Mathematical Induction and Binomial Theorem *keyboard_arrow_right*

Sequences and Series *keyboard_arrow_right*

Matrices and Determinants *keyboard_arrow_right*

Vector Algebra and 3D Geometry *keyboard_arrow_right*

Probability *keyboard_arrow_right*

Statistics *keyboard_arrow_right*

Mathematical Reasoning *keyboard_arrow_right*

Functions *keyboard_arrow_right*

Limits, Continuity and Differentiability *keyboard_arrow_right*

Differentiation *keyboard_arrow_right*

Application of Derivatives *keyboard_arrow_right*

Indefinite Integrals *keyboard_arrow_right*

Definite Integrals and Applications of Integrals *keyboard_arrow_right*

Differential Equations *keyboard_arrow_right*

Straight Lines and Pair of Straight Lines *keyboard_arrow_right*

Circle *keyboard_arrow_right*

Conic Sections *keyboard_arrow_right*