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1

### 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

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

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

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

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

2

### 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).

3

### WB JEE 2009

If C is a point on the line segment joining A($$-$$3, 4) and B(2, 1) such that AC = 2BC, then the coordinate of C is

A
$$\left( {{1 \over 3},2} \right)$$
B
$$\left( {2,{1 \over 3}} \right)$$
C
(2, 7)
D
(7, 2)

## Explanation

A($$-$$3, 4) and B(2, 1)

AC = 2BC

or, $${{AC} \over {BC}} = 2:1$$

Using section formula,

$$h = {{2\,.\,2 + 1\,.\,( - 3)} \over {2 + 1}} = {1 \over 3}$$

$$k = {{2\,.\,1 + 1\,.\,4} \over {2 + 1}} = 2$$

$$\therefore$$ Coordinates of C are $$\left( {{1 \over 3},2} \right)$$

4

### WB JEE 2008

The co-ordinates of the foot of perpendicular from (a, 0) on the line $$y = mx + {a \over m}$$ are

A
$$\left( {0,{a \over m}} \right)$$
B
$$\left( {0, - {a \over m}} \right)$$
C
$$\left( {{a \over m},0} \right)$$
D
$$\left( { - {a \over m},0} \right)$$

## Explanation

Slope of given line = m

Equation of line is $$y = mx + {a \over m}$$

Slope of perpendicular line $$PQ = {{ - 1} \over m}$$

Equation of perpendicular line is

$$y - 0 = {{ - 1} \over m}(x - a)$$ or $$my = - x + a$$

$$y = - {1 \over m}x + {a \over m}$$ ..... (i)

$$y = mx + {a \over m}$$ ..... (ii)

Subtracting (i) from (ii)

$$\left( {m + {1 \over m}} \right)x = 0$$

Putting x = 0 in (i), $$y = {a \over m}$$

So point $$Q\left( {0,{a \over m}} \right)$$.

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