AIEEE 2004 | Straight Lines and Pair of Straight Lines Question 139 | Mathematics

AIEEE 2004

MCQ (Single Correct Answer)

-1

The equation of the straight line passing through the point $$(4, 3)$$ and making intercepts on the co-ordinate axes whose sum is $$-1$$ is :

$${x \over 2} - {y \over 3} = 1$$ and $${x \over -2} +{y \over 1} = 1$$

$${x \over 2} - {y \over 3} = -1$$ and $${x \over -2} +{y \over 1} = -1$$

$${x \over 2} + {y \over 3} = 1$$ and $${x \over 2} +{y \over 1} = 1$$

$${x \over 2} + {y \over 3} = -1$$ and $${x \over -2} +{y \over 1} = -1$$

AIEEE 2003

MCQ (Single Correct Answer)

-1

Out of Syllabus

If the pair of straight lines $${x^2} - 2pxy - {y^2} = 0$$ and $${x^2} - 2qxy - {y^2} = 0$$ be such that each pair bisects the angle between the other pair, then :

$$pq = -1$$

$$p = q$$

$$p = -q$$

$$pq = 1$$.

AIEEE 2003

MCQ (Single Correct Answer)

-1

A square of side a lies above the $$x$$-axis and has one vertex at the origin. The side passing through the origin makes an angle $$\alpha \left( {0 < \alpha < {\pi \over 4}} \right)$$ with the positive direction of x-axis. The equation of its diagonal not passing through the origin is :

$$y\left( {\cos \alpha + \sin \alpha } \right) + x\left( {\cos \alpha - \sin \alpha } \right) = a$$

$$y\left( {\cos \alpha - \sin \alpha } \right) - x\left( {\sin \alpha - \cos \alpha } \right) = a$$

$$y\left( {\cos \alpha + \sin \alpha } \right) + x\left( {\sin \alpha - \cos \alpha } \right) = a$$

$$y\left( {\cos \alpha + \sin \alpha } \right) + x\left( {\sin \alpha + \cos \alpha } \right) = a$$

AIEEE 2003

MCQ (Single Correct Answer)

-1

If the equation of the locus of a point equidistant from the point $$\left( {{a_{1,}}{b_1}} \right)$$ and $$\left( {{a_{2,}}{b_2}} \right)$$ is
$$\left( {{a_1} - {a_2}} \right)x + \left( {{b_1} - {b_2}} \right)y + c = 0$$ , then the value of $$'c'$$ is :

$$\sqrt {{a_1}^2 + {b_1}^2 - {a_2}^2 - {b_2}^2} $$

$${1 \over 2}\left( {{a_2}^2 + {b_2}^2 - {a_1}^2 - {b_1}^2} \right)$$

$${{a_1}^2 - {a_2}^2 + {b_1}^2 - {b_2}^2}$$

$${1 \over 2}\left( {{a_1}^2 + {a_2}^2 + {b_1}^2 + {b_2}^2} \right)$$.

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