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

If a variable line drawn through the intersection of the lines $${x \over 3} + {y \over 4} = 1$$ and $${x \over 4} + {y \over 3} = 1,$$ meets the coordinate axes at A and B, (A $$ \ne $$ B), then the locus of the midpoint of AB is :

A

6xy = 7(x + y)

B

4(x + y)^{2} − 28(x + y) + 49 = 0

C

7xy = 6(x + y)

D

14(x + y)^{2} − 97(x + y) + 168 = 0

L_{1} : 4x + 3y $$-$$ 12 = 0

L_{2} : 3x + 4y $$-$$ 12 = 0

Equation of line passing through the intersection of these two lines L_{1} and L_{2} is

L_{1} + $$\lambda $$L_{2} = 0

$$ \Rightarrow $$$$\,\,\,$$(4x + 3y $$-$$ 12) + $$\lambda $$(3x + 4y $$-$$ 12) = 0

$$ \Rightarrow $$$$\,\,\,$$ x(4 + 3$$\lambda $$) + y(3 + 4$$\lambda $$) $$-$$ 12(1 + $$\lambda $$) = 0

this line meets x coordinate at point A and y coordinate at point B.

$$\therefore\,\,\,$$ Point A = $$\left( {{{12\left( {1 + \lambda } \right)} \over {4 + 3\lambda }},0} \right)$$

and Point B = $$\left( {0,\,\,{{12\left( {1 + \lambda } \right)} \over {3 + 4\lambda }}} \right)$$

Let coordinate of midpoint of line AB is (h, k).

$$\therefore\,\,\,$$ h = $${{6\left( {1 + \lambda } \right)} \over {4 + 3\lambda }}$$ . . . . . (1)

and k = $${{6\left( {1 + \lambda } \right)} \over {3 + 4\lambda }}$$ . . . . (2)

Eliminate $$\lambda $$ from (1) and (2), then we get

6(h + k) = 7 hk

$$\therefore\,\,\,$$ Locus of midpoint of line AB is ,

6(x + y) = 7xy

L

Equation of line passing through the intersection of these two lines L

L

$$ \Rightarrow $$$$\,\,\,$$(4x + 3y $$-$$ 12) + $$\lambda $$(3x + 4y $$-$$ 12) = 0

$$ \Rightarrow $$$$\,\,\,$$ x(4 + 3$$\lambda $$) + y(3 + 4$$\lambda $$) $$-$$ 12(1 + $$\lambda $$) = 0

this line meets x coordinate at point A and y coordinate at point B.

$$\therefore\,\,\,$$ Point A = $$\left( {{{12\left( {1 + \lambda } \right)} \over {4 + 3\lambda }},0} \right)$$

and Point B = $$\left( {0,\,\,{{12\left( {1 + \lambda } \right)} \over {3 + 4\lambda }}} \right)$$

Let coordinate of midpoint of line AB is (h, k).

$$\therefore\,\,\,$$ h = $${{6\left( {1 + \lambda } \right)} \over {4 + 3\lambda }}$$ . . . . . (1)

and k = $${{6\left( {1 + \lambda } \right)} \over {3 + 4\lambda }}$$ . . . . (2)

Eliminate $$\lambda $$ from (1) and (2), then we get

6(h + k) = 7 hk

$$\therefore\,\,\,$$ Locus of midpoint of line AB is ,

6(x + y) = 7xy

2

The point (2, 1) is translated parallel to the line L : x− y = 4 by $$2\sqrt 3 $$ units. If the newpoint Q lies in the third quadrant, then the equation of the line passing through Q and perpendicular to L is :

A

x + y = 2 $$-$$ $$\sqrt 6 $$

B

x + y = 3 $$-$$ 3$$\sqrt 6 $$

C

x + y = 3 $$-$$ 2$$\sqrt 6 $$

D

2x + 2y = 1 $$-$$ $$\sqrt 6 $$

x $$-$$ y = 4

To find equation of R

slope of L = 0 is 1

$$ \Rightarrow $$ slope of QR = $$-$$ 1

Let QR is y = mx + c

y = $$-$$ x + c

x + y $$-$$ c = 0

distance of QR from (2, 1) is 2$$\sqrt 3 $$

2$$\sqrt 3 $$ = $${{\left| {2 + 1 - c} \right|} \over {\sqrt 2 }}$$

2$$\sqrt 6 $$ = $$\left| {3 - c} \right|$$

c $$-$$ 3 = $$ \pm 2\sqrt 6 $$ c = 3 $$ \pm $$ 2$$\sqrt 6 $$

Line can be x + y = 3 $$ \pm $$ 2$$\sqrt 6 $$

x + y = 3 $$-$$ 2$$\sqrt 6 $$

To find equation of R

slope of L = 0 is 1

$$ \Rightarrow $$ slope of QR = $$-$$ 1

Let QR is y = mx + c

y = $$-$$ x + c

x + y $$-$$ c = 0

distance of QR from (2, 1) is 2$$\sqrt 3 $$

2$$\sqrt 3 $$ = $${{\left| {2 + 1 - c} \right|} \over {\sqrt 2 }}$$

2$$\sqrt 6 $$ = $$\left| {3 - c} \right|$$

c $$-$$ 3 = $$ \pm 2\sqrt 6 $$ c = 3 $$ \pm $$ 2$$\sqrt 6 $$

Line can be x + y = 3 $$ \pm $$ 2$$\sqrt 6 $$

x + y = 3 $$-$$ 2$$\sqrt 6 $$

3

A ray of light is incident along a line which meets another line, 7x − y + 1 = 0, at the point (0, 1). The ray is then reflected from this point along the line, y + 2x = 1. Then the equation of the line of incidence of the ray of light is :

A

41x − 38y + 38 = 0

B

41x + 25y − 25 = 0

C

41x + 38y − 38 = 0

D

41x − 25y + 25 = 0

Let slope of incident ray be m.

$$ \therefore $$ angle of incidence = angle of reflection

$$ \therefore $$ $$\left| {{{m - 7} \over {1 + 7m}}} \right| = \left| {{{ - 2 - 7} \over {1 - 14}}} \right| = {9 \over {13}}$$

$$ \Rightarrow $$ $${{m - 7} \over {1 + 7m}} = {9 \over {13}}$$

or $${{m - 7} \over {1 + 7m}} = - {9 \over {13}}$$

$$ \Rightarrow $$ 13m $$-$$ 91 $$=$$ 9 + 63m

or 13m $$-$$ 91 $$=$$ $$-$$ 9 $$-$$ 63m

$$ \Rightarrow $$ 50m $$=$$ $$-$$ 100 or 76m $$=$$ 82

$$ \Rightarrow $$ m $$=$$ $$ - {1 \over 2}$$

or m $$=$$ $${{41} \over {38}}$$

$$ \Rightarrow $$ y $$-$$ 1 $$=$$ $$-$$ $${1 \over 2}$$ (x $$-$$ 0)

or y $$-$$ 1 $$=$$ $${{41} \over {38}}$$ (x $$-$$ 0)

i.e x + 2y $$-$$ 2 $$=$$ 0

or 38y $$-$$ 38 $$-$$ 41x $$=$$ 0

$$ \Rightarrow $$ 41x $$-$$ 38y + 38 $$=$$ 0

$$ \therefore $$ angle of incidence = angle of reflection

$$ \therefore $$ $$\left| {{{m - 7} \over {1 + 7m}}} \right| = \left| {{{ - 2 - 7} \over {1 - 14}}} \right| = {9 \over {13}}$$

$$ \Rightarrow $$ $${{m - 7} \over {1 + 7m}} = {9 \over {13}}$$

or $${{m - 7} \over {1 + 7m}} = - {9 \over {13}}$$

$$ \Rightarrow $$ 13m $$-$$ 91 $$=$$ 9 + 63m

or 13m $$-$$ 91 $$=$$ $$-$$ 9 $$-$$ 63m

$$ \Rightarrow $$ 50m $$=$$ $$-$$ 100 or 76m $$=$$ 82

$$ \Rightarrow $$ m $$=$$ $$ - {1 \over 2}$$

or m $$=$$ $${{41} \over {38}}$$

$$ \Rightarrow $$ y $$-$$ 1 $$=$$ $$-$$ $${1 \over 2}$$ (x $$-$$ 0)

or y $$-$$ 1 $$=$$ $${{41} \over {38}}$$ (x $$-$$ 0)

i.e x + 2y $$-$$ 2 $$=$$ 0

or 38y $$-$$ 38 $$-$$ 41x $$=$$ 0

$$ \Rightarrow $$ 41x $$-$$ 38y + 38 $$=$$ 0

4

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

Number in Brackets after Paper Name Indicates No of Questions

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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*

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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*