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JEE Mathematics Solved Question Paper 2014

Multiple Choice Questions

The equation of a straight line; perpendicular to 8x - 4y = 6 and forming a triangle of area 6sq. units with coordinate axes, is

x - 2y = 6
4x + 3y = 12
4x + 3y + 24 = 0
3x + 4y = 12

If the image of $(\frac{- 7}{6}, \frac{- 6}{5}) in a line is (1, 2), then the equation of the line is$

4x + 3y = 1
3x - y = 0
4x - y = 0
3x + 4y = 1

If a line l passes through (k, 2k), (3k, 3k) and (3, 1), k $\neq$ 0, then the distance from the origin to the line is

$\frac{1}{\sqrt{5}}$
$\frac{4}{\sqrt{5}}$
$\frac{3}{\sqrt{5}}$
$\frac{2}{\sqrt{5}}$

4.
The area (in sq. units) of the triangle formed by the lines x² - 3y + y = 0 and x + y + 1 = 0, is

$\frac{2}{\sqrt{3}}$

$\frac{\sqrt{3}}{2}$

$5 \sqrt{2}$

$\frac{1}{2 \sqrt{5}}$

$\frac{1}{2 \sqrt{5}}$

$Given equations of line are x^{2} - 3 xy + y^{2} = 0 and x + y + 1 = 0 Let, m_{1} and m_{2} be the slope of the line x^{2} - 3 xy + y^{2} = 2 ∴ m_{1} + m_{2} = - \frac{coefficient of xy}{Coefficient of y^{2}} = + \frac{3}{1} = 3 . . . (i) and m_{1} m_{2} = \frac{Coefficient of x^{2}}{Coefficient of y^{2}} = \frac{1}{1} = 1 . . . (ii) Now, m_{1} - m_{2} = \sqrt{{(m_{1} + m_{2})}^{2} - 4 m_{1} m_{2}} = \sqrt{{(3)}^{2} - 4 \times 1} = \sqrt{9 - 4} = \sqrt{5} \Rightarrow m_{1} - m_{2} = \sqrt{5} . . . (iii)$

On solving eqs. (i) and (iii), we get

$m_{1} = \frac{3 + \sqrt{5}}{2} and m_{2} = \frac{3 - \sqrt{5}}{2} ∴ Equation of lines will be y = \frac{3 + \sqrt{5}}{2} x . . . (iv) and y = \frac{3 - \sqrt{5}}{2} x . . . (v) and third line is given x + y + 1 = 0 . . . (vi) ∴ The points of intersection of these lines are A (0, 0), B (- \frac{2}{5 + \sqrt{5}}, \frac{3 + \sqrt{5}}{5 + \sqrt{5}}), and C (- \frac{2}{5 - \sqrt{5}}, \frac{3 - \sqrt{5}}{5 - \sqrt{5}}) ∴ Area of triangle = \frac{1}{2} |\begin{matrix} 0 & 0 & 1 \\ - \frac{2}{5 + \sqrt{5}} & \frac{3 + \sqrt{5}}{5 + \sqrt{5}} & 1 \\ - \frac{2}{5 - \sqrt{5}} & \frac{3 - \sqrt{5}}{5 - \sqrt{5}} & 1 \end{matrix}| = \frac{1}{2} [0 + 0 + 1 (\frac{- 2 (3 - \sqrt{5})}{5^{2} - {(\sqrt{5})}^{2}} + \frac{2 (3 + \sqrt{5})}{5^{2} - {(\sqrt{5})}^{2}})] = \frac{1}{2} [\frac{4 \sqrt{5}}{20}] = \frac{1}{2 \sqrt{5}}$