MCA Entrance Previous Year Question Bank

The product of perpendiculars from $(2,-1)$ to the pair of lines $2x^2+6xy+y^2=0$ is:

The equation of line passing through $(-2,3)$ and perpendicular to $2x-3y+6=0$ is given by:

The distance between the pair of straight lines $x^2+2\sqrt{3}xy+2y^2+4x+1=0$ is:

If $y=mx$ bisects the angle between the lines $x^2({\tan }^2\theta +{\cos }^2\theta )+2xy\tan \theta -y^2{\sin }^2\theta =0$ when $\theta =\frac{\pi }{3}$, then the value of $\sqrt{3}{m}^2+4m$ is:

If $(-4,5)$ is one vertex and $7x-y+8=0$ is one diagonal of a square, then the equation of the other diagonal is:

The foot of the perpendicular from the point $(2,4)$ upon $x+y=1$ is:

If the two pairs of lines $x^2-2mxy-y^2=0$ and $x^2-2nxy-y^2=0$ are such that one of them represents the bisector of the angles between the other, then:

The equation of the line parallel to the line $2x-3y=7$ and passing through the middle point of the line segment joining the points $(1,3)$ and $(1,-7)$ is:

If the straight line $ax+by+c=0$ always passes through $(1,-2)$, then $a,b,c$ are in:

If $2x^2+7xy+3y^2+8x+14y+\lambda =0$ represents a pair of straight lines, the value of $\lambda$ is