One possible condition for the three points (a, b), (b, a) and (a², $$-$$ b²) to be collinear is

The distance between the lines $$5x - 12y + 65 = 0$$ and $$5x - 12y - 39 = 0$$ is

The co-ordinates of the foot of perpendicular from (a, 0) on the line $$y = mx + {a \over m}$$ are

If C is a point on the line segment joining A($$-$$3, 4) and B(2, 1) such that AC = 2BC, then the coordinate of C is

The coordinates of the foot of the perpendicular from (0, 0) upon the line x + y = 2 are

A line through the point A(2, 0) which makes an angle of 30$$^\circ$$ with the positive direction of x-axis is rotated about A in clockwise direction through an angle 15$$^\circ$$. Then the equation of the straight line in the new position is

If C is the reflection of A(2, 4) in x-axis and B is the reflection of C in y-axis, then $$|AB|$$ is

The number of points on the line x + y = 4 which are unit distance apart from the line 2x + 2y = 5 is

The point ($$-$$4, 5) is the vertex of a square and one of its diagonals is 7x $$-$$ y + 8 = 0. The equation of the other diagonal is

The straight line 3x + y = 9 divides the line segment joining the points (1, 3) and (2, 7) in the ratio

If the sum of distances from a point P on two mutually perpendicular straight lines is 1 unit, then the locus of P is a/an

If the three points (3q, 0) (0, 3p) and (1, 1) are collinear then which one is true?

The equations $$y = \pm \sqrt {3x} $$, y = 1 are the sides of

The equations of the lines through (1, 1) and making angles of 45$$^\circ$$ with the line x + y = 0 are

The coordinates of the two points lying on x + y = 4 and at a unit distance from the straight line 4x + 3y = 10 are

If the three points A(1, 6), B(3, $$-$$4) and C(x, y) are collinear then the equation satisfying by x and y is

The equation of the locus of the point of intersection of the straight lines $$x\sin \theta + (1 - \cos \theta )y = a\sin \theta $$ and $$x\sin \theta - (1 + \cos \theta )y + a\sin \theta = 0$$ is

Consider three points $P(\cos \alpha, \sin \beta), Q(\sin \alpha, \cos \beta)$ and $R(0,0)$, where $0<\alpha, \beta<\frac{\pi}{4}$. Then

The line parallel to the $x$-axis passing through the intersection of the lines $a x+2 b y+3 b=0$ and $b x-2 a y-3 a=0$ where $(a, b) \neq(0,0)$ is

If $$(1,5)$$ be the midpoint of the segment of a line between the line $$5 x-y-4=0$$ and $$3 x+4 y-4=0$$, then the equation of the line will be

In $$\triangle \mathrm{ABC}$$, co-ordinates of $$\mathrm{A}$$ are $$(1,2)$$ and the equation of the medians through $$\mathrm{B}$$ and C are $$x+\mathrm{y}=5$$ and $$x=4$$ respectively. Then the midpoint of $$\mathrm{BC}$$ is

A, B are fixed points with coordinates (0, a) and (0, b) (a > 0, b > 0). P is variable point (x, 0) referred to rectangular axis. If the angle $$\angle$$APB is maximum, then

The equation $${r^2}{\cos ^2}\left( {\theta - {\pi \over 3}} \right) = 2$$ represents

If $$4{a^2} + 9{b^2} - {c^2} + 12ab = 0$$, then the family of straight lines $$ax + by + c = 0$$ is concurrent at

The straight lines $$x + 2y - 9 = 0,3x + 5y - 5 = 0$$ and $$ax + by - 1 = 0$$ are concurrent if the straight line $$35x - 22y + 1 = 0$$ passes through the point

The locus of points (x, y) in the plane satisfying $${\sin ^2}x + {\sin ^2}y = 1$$ consists of

If the algebraic sum of the distances from the points (2, 0), (0, 2) and (1, 1) to a variable straight line be zero, then the line passes through the fixed point

If the sum of the distances of a point from two perpendicular lines in a plane is 1 unit, then its locus is

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.

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.

Find the image of the point ($$-$$8, 12) with respect to the line 4x + 7y + 13 = 0.

Let $$\Gamma$$ be the curve $$\mathrm{y}=\mathrm{be}^{-x / a}$$ & $$\mathrm{L}$$ be the straight line $$\frac{x}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1$$ where $$\mathrm{a}, \mathrm{b} \in \mathbb{R}$$. Then

A square with each side equal to '$$a$$' above the $$x$$-axis and has one vertex at the origin. One of the sides passing through the origin makes an angle $$\alpha$$ $$\left(0<\alpha< \frac{\pi}{4}\right)$$ with the positive direction of the axis. Equation of the diagonals of the square

If $$\mathrm{ABC}$$ is an isosceles triangle and the coordinates of the base points are $$B(1,3)$$ and $$C(-2,7)$$. The coordinates of $$A$$ can be

A rectangle ABCD has its side parallel to the line y = 2x and vertices A, B, D are on lines y = 1, x = 1 and x = $$-$$1 respectively. The coordinate of C can be

Consider the equation $$y - {y_1} = m(x - {x_1})$$. If m and x₁ are fixed and different lines are drawn for different values of y₁, then