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A line intersects x and y–axes at P and Q respectively. If (2, –5) is the mid–point of PQ, then the coordinates of P and Q are respectively.
Mid point of and
is
A/Q
Coordinate of P and Q are (0, -10) and (4, 0)
The distance between the points \((\cos\theta,\sin\theta) \) and \((\sin\theta,-\cos\theta) \) is:
If the points (1, 2), (–5, 6) and (a, – 2) are collinear, then a is equal to :
The points A(9, 0), B(9, 6), C(– 9, 6) and D(– 9, 0) are the vertices of a :
and
opposite sides are equal so it is a rectangle
AOBC is a rectangle whose three vertices are A(0, 3), O(0, 0) and B(5, 0). The length of its diagonal is :
The points (– 4, 0), (4, 0) and (0, 3) are vertices of a:
It is an isosceles triangle
The coordinates of the vertices of an equilateral triangle are \(A(3, y), B (3,\sqrt3) \) and \(C(0,0) \). The value of y is :
The distance between the points \((a\cos\theta+b\sin\theta,0) \) and \((0,a\sin\theta-b\cos\theta) \) is:
(–1, 2), (2, –1) and (3, 1) are three vertices of a parallelogram. The coordinates of the fourth vertex are :
Let the fourth vertex be a,b
Mid point of
The fourth vertex = (0, 4)
If A(5, 3), B(11, –5) and C(12, a) are the vertices of a right angled triangle, right angled at C, then the value of a is :
So,
The perimeter of the triangle with vertices (0, 4),(0, 0) and (3, 0) is :
Perimeter
The point which lies on the perpendicular bisector of the line segment joining the points A(– 2, –5) and B(2, 5) is :
The area of the triangle with vertices \((a, b + c),(b, c + a)\) and \((c, a + b)\) is :
If the centroid of the triangle formed by the points \((a, b), (b, c)\) and \((c, a)\) is at the origin, then \(a^3 + b^3 + c^3\) is equal to :
The line segment joining points (–3, – 4) and (1, – 2) is divided by y-axis in the ratio :
If points \((x,0),\ (0,y) \) are \((1,1) \) and collinear, then \(\dfrac{1}{x}+\dfrac{1}{y} \) is equal to:
If the ratio in which P divides the line segment joining \((x_1,y_1) \) and \((x_2,y_2) \) be k : 1, then the coordinates of the point P are :
Area of a triangle is taken :
Area of a triangle is taken always positive
The area of the triangle formed by \((x, y + z),(y, z + x)\) and \((z, x + y)\) is :
The fourth vertex D of a parallelogram ABCD whose three vertices are A(–2, 3), B(6, 7) and C(8, 3) is :
The fourth vertex = (0,-1)
If the points A(1, 2), B(0, 0) and C(a, b) are collinear, then :
If are collinear
In the figure, OAB is a triangle. The coordinates of the point which is equidistant from the three vertices are :
Let the points be
Points are
A circle drawn with origin as the centre passes through \(\left(\dfrac{13}{2},0\right) \) The point which does not lie in the interior of the circle is :
(a) Distance between and
(b) Distance between and
(c) Distance between and
is
So does not be in the interior of the circle
The coordinates of the centroid of the triangle with vertices (a, 0), (0, b) and (a, b) are :
Area of the triangle with vertices (x, 0), (0, y) and (x, y) is :
If the area of a quadrilateral ABCD is zero, then the four points A, B, C and D are :
The four points will be collinear it is because whenever all points are collinear the area of any polygon is 0.
If the points (0, 0), (1, 2) and (x, y) are collinear then :
The perpendicular distance of A (5, 12) from the y-axis is :
The point of the y-axis is (0, 12)
Distance between (5, 12) and (0, 12)
The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is :
Perimeter