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The diagonal passing through origin of a quadrilateral formed by $x=0,\ y=0,\ x+y=1$ and $6x+y=3,$ is
The vertices of a triangle OBC are $(0,\ 0),\ (-3,\ -1)$ and $(-1,\ -3)\ $ respectively. Then the equation of line parallel to BC which is at $\frac{1}{2}$ unit distant from origin and cuts OB and OC, is
A vertex of square is (3, 4) and diagonal $x+2y=1,$ then the second diagonal which passes through given vertex will be
A vertex of equilateral triangle is (2, 3) and equation of opposite side is $x+y=2,$ then the equation of one side from rest two, is
A straight line moves so that the sum of the reciprocals of its intercepts on two perpendicular lines is constant, then the line passes through
If a, b, c are in harmonic progression, then straight line $\frac{x}{a}+\frac{y}{b}+\frac{1}{c}=0$ always passes through a fixed point, that point is
If the straight line $ax+by+c=0$ always passes through (1, – 2), then a, b, c are
If $u={{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0,$ $v={{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0$ and $\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{{{c}_{1}}}{{{c}_{2}}},$ then the curve $u+kv=0$ is
For what values of a and b the intercepts cut off on the coordinate axes by the line $ax+by+8=0$ are equal in length but opposite in signs to those cut off by the line $2x-3y+6=0$ on the axes
If a and b are two arbitrary constants, then the straight line $(a-2b)x+(a+3b)y+3a+4b=0$ will pass through

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