JEE Advanced Level Test: Straight Line- 4 - EmSAT Achieve MCQ

A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) are three non-collinear points in cartesian plane. Number of parallelograms that can be drawn with these three points as vertices are

Three vertices of triangle ABC are A(-1, 11), B(–9, –8) and C(15, -2). The equation of angle bisector of angle A is

If line y – x + 2 = 0 is shifted parallel to itself towards the positive direction of the x-axis by a perpendicular distance of 3√2units, then the equation of the new line is

If the axes are rotated through an angle of 30º in the anti-clockwise direction, the coordinates of point (4, –2√3) with respect to new axes are

Keeping the origin constant axes are rotated at an angle 30º in clockwise direction then new coordinate of (2, 1) with respect to old axes is

If one diagonal of a square is along the line x = 2y and one of its vertex is (3, 0), then its sides through this vertex are given by the equations

The equation 2x² + 4xy – py² + 4x + qy + 1 = 0 will represent two mutually perpendicular straight lines, if

One of the diameter of the circle circumscribing the rectangle ABCD is 4y = x + 7. If A and B are the points (–3, 4) and (5, 4) respectively then the area of rectangle is equal to

The co-ordinates of a point P on the line 2x–y+5=0 such that |PA – PB| is maximum where A is (4, –2) and B is (2, –4) will be

The line x + y = p meets the axis of x and y at A and B respectively. A triangle APQ is inscribed in the triangle OAB, O being the origin, with right angle at Q, P and Q lie respectively on OB and AB. If the area of the triangle APQ is 3/8^th of the area of the triangle OAB, then AQ/BQ is equal to

Lines, L₁ : x + √3y = 2, and L₂ : ax + by = 1, meet at P and enclose an angle of 45º between them. Line L₃ : y = , also passes through P then

A triangle is formed by the lines 2x – 3y – 6 = 0 ; 3x – y + 3 = 0 and 3x + 4y – 12 = 0. If the points P(a, 0) and Q(0, b) always lie on or inside the DABC, then

The line x + 3y – 2 = 0 bisects the angle between a pair of straight lines of which one has equation x – 7y + 5 = 0. The equation of the other line is

A ray of light passing through the point A(1, 2) is reflected at a point B on the x-axis and then passes through (5, 3). Then the equation of AB is

Let the algebraic sum of the perpendicular distances from the point (3, 0), (0, 3) & (2, 2) to a variable straight line be zero, then the line passes through a fixed point whose co-ordinates are

The pair of straight lines x² – 4xy + y² = 0 together with the line x + y + = 0 form a triangle which is

Let Aº (3, 2) and Bº (5, 1). ABP is an equilateral triangle is constructed on the side of AB remote from the origin then the orthocentre of triangle ABP is

The line PQ whose equation is x – y = 2 cuts the x-axis at P and Q is (4, 2). The line PQ is rotated about P through 45º in the anticlockwise direction. The equation of the line PQ in the new position is

Distance between two lines represented by the line pair, x² – 4xy + 4y² + x – 2y – 6 = 0 is

The circumcentre of the triangle formed by the lines, xy + 2x + 2y + 4 = 0 and x + y + 2 = 0 is

Area of the rhombus bounded by the four lines, ax ± by ± c = 0 is

If the lines ax + y + 1 = 0, x + by + 1 = 0 & x + y + c = 0 where a, b & c are distinct real numbers different from 1 are concurrent, then the value of  +  +  equals

The area enclosed by 2 | x | + 3| y | < 6 is

The point (4, 1) undergoes the following three transformations successively
(i) Reflection about the line y = x
(ii) Translation through a distance 2 units along the positive direction of x-axis
(iii) Rotation through an angle p/4 about the origin in the counter clockwise direction.
The final position of the points is given by the coordinates