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1

### WB JEE 2009

The number of points on the line x + y = 4 which are unit distance apart from the line 2x + 2y = 5 is

A
0
B
1
C
2
D
infinity

## Explanation

Since x + y = 4 and 2x + 2y = 5 are parallel.

Take (4, 0) on the line x + y = 4

Distance of (4, 0) from the line 2x + 2y $$-$$ 5 = 0 is

$${{|2\,.\,4 + 2\,.\,0 - 5|} \over {\sqrt {{2^2} + {2^2}} }} = {3 \over {2\sqrt 2 }} = {{3\sqrt 2 } \over 4} > 1$$ (unit distance)

$$\because$$ Both lines are parallel and at a distance greater than unity

$$\therefore$$ There is no point on the line x + y = 4.

2

### WB JEE 2009

If C is the reflection of A(2, 4) in x-axis and B is the reflection of C in y-axis, then $$|AB|$$ is

A
20
B
2$$\sqrt5$$
C
4$$\sqrt5$$
D
4

## Explanation

Reflection of point (h, k) about x-axis is (h, $$-$$k)

Reflection of point (h, k) about y-axis is ($$-$$h, k)

$$\therefore$$ Reflection point C of A(2, 4) in x-axis is (2, $$-$$4) and reflection point B of C(2, $$-$$4) in y-axis is ($$-$$2, $$-$$4)

$$\therefore$$ $$|AB| = \sqrt {{{(2 + 2)}^2} + {{(4 + 4)}^2}} = 4\sqrt 5$$

3

### WB JEE 2009

A line through the point A(2, 0) which makes an angle of 30$$^\circ$$ with the positive direction of x-axis is rotated about A in clockwise direction through an angle 15$$^\circ$$. Then the equation of the straight line in the new position is

A
$$(2 - \sqrt 3 )x + y - 4 + 2\sqrt 3 = 0$$
B
$$(2 - \sqrt 3 )x - y - 4 + 2\sqrt 3 = 0$$
C
$$(2 - \sqrt 3 )x - y + 4 + 2\sqrt 3 = 0$$
D
$$(2 - \sqrt 3 )x + y + 4 + 2\sqrt 3 = 0$$

## Explanation

Equation of line which is passing through (x1, y1) and angle of inclination $$\theta$$ with +ve x axis in anti-clockwise is

$$y - {y_1} = \tan \theta (x - {x_1})$$

$$\therefore$$ $$(y - 0) = \tan 15^\circ (x - 2)$$

$$[\tan 15^\circ = \tan (45^\circ - 30^\circ ) = {{\tan 45^\circ - \tan 30^\circ } \over {1 + \tan 45^\circ \tan 30^\circ }} = {{\sqrt 3 - 1} \over {\sqrt 3 + 1}} = 2 - \sqrt 3 ]$$

$$\Rightarrow (2 - \sqrt 3 )x - y - 4 + 2\sqrt 3 = 0$$

4

### WB JEE 2009

The coordinates of the foot of the perpendicular from (0, 0) upon the line x + y = 2 are

A
(2, $$-$$1)
B
($$-$$2, 1)
C
(1, 1)
D
(1, 2)

## Explanation

Let coordinates of foot P are (h, k)

$$\therefore$$ Slope of OP = k/h

Slope of given line = $$-$$1

$$\because$$ LM $$\bot$$ OP

$$\therefore$$ $${k \over h} \times - 1 = - 1$$

$$\Rightarrow$$ k = h ..... (i)

Also point P lies on the given line

$$\therefore$$ k + h = 2 ..... (ii)

Solving (i) and (ii), we get h = 1, k = 1

So coordinates of foot are (1, 1).

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