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1

### WB JEE 2009

MCQ (Single Correct Answer)

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 $$\left( {{1 \over 3},2} \right)$$

2

### WB JEE 2008

MCQ (Single Correct Answer)

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

3

### WB JEE 2008

MCQ (Single Correct Answer)

The distance between the lines $$5x - 12y + 65 = 0$$ and $$5x - 12y - 39 = 0$$ is

A
4
B
16
C
2
D
8

## Explanation

Given equations of lines are $$5x - 12y + 65 = 0$$ and $$5x - 12y - 39 = 0$$

Since given lines are parallel

$$\therefore$$ Perpendicular distance $$= {{|65 + 39|} \over {\sqrt {{{(5)}^2} + {{(12)}^2}} }}$$

$$= {{104} \over {\sqrt {25 + 144} }} = {{104} \over {13}} = 8$$.

4

### WB JEE 2008

MCQ (Single Correct Answer)

One possible condition for the three points (a, b), (b, a) and (a2, $$-$$ b2) to be collinear is

A
a $$-$$ b = 2
B
a + b = 2
C
a = 1 + b
D
a = 1 $$-$$ b

## Explanation

Let A(a, b), B(b, a), C(a2, $$-$$ b2) be three points which are collinear.

$$\therefore$$ $$\left| {\matrix{ a & b & 1 \cr b & a & 1 \cr {{a^2}} & { - {b^2}} & 1 \cr } } \right| = 0$$

$$\Rightarrow 1( - {b^3} - {a^3}) - 1( - a{b^2} - {a^2}b) + 1({a^2} - {b^2}) = 0$$

$$\Rightarrow - ({a^3} + {b^3}) + ab(a + b) + (a + b)(a - b) = 0$$

$$= - (a + b)({a^2} + {b^2} - ab) + ab(a + b) + (a + b)(a - b) = 0$$

$$\Rightarrow (a + b)( - {a^2} - {b^2} + ab + ab + a - b) = 0$$

$$\Rightarrow (a + b)( - {a^2} - {b^2} + 2ab + a - b) = 0$$

$$\Rightarrow (a + b)( - ({a^2} + {b^2} - 2ab) + (a - b)) = 0$$

$$\Rightarrow (a + b)( - {(a - b)^2} + (a - b)) = 0$$

$$\Rightarrow (a + b)(a - b)( - (a - b) + 1) = 0$$

$$\Rightarrow ({a^2} - {b^2})(1 - a + b) = 0$$

$$\therefore$$ either $${a^2} - {b^2} = 0$$ or $$1 - a + b = 0 \Rightarrow a = 1 + b$$.

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