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1

WB JEE 2009

Subjective

The equations to the pairs of opposite sides of a parallelogram are x2 $$-$$ 5x + 6 = 0 and y2 $$-$$ 6y + 5 = 0. Find the equations of its diagonals.

Answer

.

Explanation

$${x^2} - 5x + 6 = 0 \Rightarrow (x - 3)(x - 2) = 0$$

$$\Rightarrow$$ x = 3, x = 2 are the equations of lines $${y^2} - 6y + 5 = 0$$

$$ \Rightarrow (y - 5)(y - 1) = 0$$

$$\Rightarrow$$ y = 5, y = 1 are the equations of lines ABCD is parallelogram formed by above lines

co-ordinates of A(2, 1)

co-ordinates of B(3, 1)

co-ordinates of C(3, 5)

co-ordinates of D(2, 5)

Equation of diagonal $$AC:{{{y_2} - {y_1}} \over {{x_2} - {x_1}}} = {{y - {y_1}} \over {x - {x_1}}}$$

$$ \Rightarrow {{5 - 1} \over {3 - 2}} = {{y - 1} \over {x - 2}} \Rightarrow {4 \over 1} = {{y - 1} \over {x - 2}}$$

$$ \Rightarrow 4x - 8 = y - 1 \Rightarrow 4x - y = 7$$

Equation of diagonal $$BD:{{5 - 1} \over {2 - 3}} = {{y - 1} \over {x - 3}}$$

$$ \Rightarrow {4 \over { - 1}} = {{y - 1} \over {x - 3}} \Rightarrow 4x - 12 = - y + 1$$

$$ \Rightarrow 4x + y = 13$$

$$\therefore$$ Required equations of diagonals are

4x $$-$$ y = 7 and 4x + y = 13

2

WB JEE 2008

Subjective

If 2a $$-$$ 5b $$-$$ 3c = 0, show that the straight line ax + by + c = 0 always passes through a fixed point. Find the equation of one straight line passing through this fixed point and parallel to the line 9x + 12y = 20.

Answer

.

Explanation

We have, $$2a - 5b - 3c = 0$$.

or, $$c = {{2a - 5b} \over 3}$$ ..... (i)

Hence, equation of the straight line can be written as

$$ax + by + {{2a - 5b} \over 3} = 0$$

or, $$3ax + 3by + 2a - 5b = 0$$

or, $$a(3x + 2) + b(3y - 5) = 0$$.

This will be true for all values of a and b, only if (3x + 2) and (3y $$-$$ 5) are simultaneously zero which means, for x = $$-$$2/3 and y = 5/3, the equation is always satisfied for any values of a and b.

$$\therefore$$ The straight line passes through the point ($$-$$2/3, 5/3)

Now, slope of the given line 9x + 12y = 20 is $$-$$3/4

Any straight line parallel to this line will also have slope of $$-$$3/4

$$\therefore$$ Equation of the line parallel to 9x + 12y = 20 and passing through ($$-$$2/3, 5/3) is

$$y - {5 \over 3} = - {3 \over 4}\left( {x + {2 \over 3}} \right)$$

$$y - {5 \over 3} = - {3 \over 4}\left( {x + {2 \over 3}} \right)$$ or, $$ - 4y + {{20} \over 3} = 3x + 2$$

or, $$3x + 4y - {{14} \over 3} = 0$$ or, $$9x + 12y - 14 = 0$$.

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