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The distance of the point (– 3, 4) from the origin is
Let the given point be (x_{1}, y_{1}) = (3, 4) and the orgin is (x_{2}, y_{2}) = (0, 0)
∴ Distance of the given point from the orgin =
If A and B are the points (– 6, 7) and (– 1, – 5) respectively, then the distance 2AB is equal to 26 units 13 units 15 units 20 units 2AB
The distance between the points (a, a) and (−√3a,√3a) is
Three consecutive vertices of a parallelogram ABCD are A(1, 2), B(1, 0) and C(4, 0). The co – ordinates of the fourth vertex D are
Let coordinates of D be (x, y).
Since diagonals of a parallelogram bisect each other.
Therefore, coordinates of O will be
Therefore, the required coordinates are (4, 2).
If (a, 0) , (0, b) and (x, y) are collinear, then
If given points are collinear, the triangle formed by these three points is 0.
The distance between the points (– 1, – 5) and (– 6, 7) is
Let point A be (x_{1}, y_{1}) = (1, 5) and point B (x_{2}, y_{2}) = (6, 7)
∴Distance between A and B
If P(x, y) is any point on the line joining the points A(a, 0) and B(0, b), then
The distance between the points A(p sin 25^{∘}, 0) and B(0, p sin 65^{∘}) is
p units p^{2} units 1 units 0 units Distance between
The length of the median through A of ΔABC with vertices A(7, – 3), B(5, 3) and C(3, – 1) is
If the co – ordinates of a point are (3, – 7), then its ordinate is
Since y−coordinate is called ordinate.
Therefore, ordinate is −7.
The distance between the points (a, b) and (– a, b) is
Let point A be (a, b) and point B is (a, b)
∴ distance between A and B
The vertices of a quadrilateral are (1, 7), (4, 2), (– 1, – 1) and (– 4, 4). The quadrilateral is a
Let A (1, 7), B (4, 2), C (1, 1) and D(4, 4) are the vertices of a quardinate
Since, all sides are equal and both diagonals are also equal.
Therefore, the given quadrilateral is a square.
The distance between
If the line segment joining the points A(x_{1},y_{1}) and B(x_{2},y_{2}) is divided by a point P in the ratio 1 : k internally, then the co – ordinates of the point P are
If the co – ordinates of a point are (– 5, 11), then its abscissa is
Since x−coordinate is called abscissa.
Therefore, abscissa is −5.
If the distance between the points (p, – 5) and (2, 7) is 13 units, then the value of ‘p’ is
Let point A be (p,−5)(p,−5) and point B (2, 7) and distance between A and B = 13 units
If A is point on the x – axis whose abscissa is 5 and B is the point (1, – 3), then the distance AB is
The point where the medians of a triangle meet is called the ________ of the triangle
The point where the medians of a triangle meet is called thecentroidof the triangle.
The base of an equilateral triangle ABC lies on the y – axis. The co – ordinates of the point C is (0, – 3). If origin is the midpoint of BC, then the co – ordinates of B are
Since coordinates of any point on y−axis is (0,y).
Therefore, abscissa is 0.
The values of ‘y’ for which the distance between the points (2, – 3) and (10, y) is 10 units is
Let one point be A (2,−3) and 2nd point is B (10,y) and
distance between point A and point B = AB = 10 units
The distance of the point (– 5, 12) from the y – axis is
The distance of any point from y−axis is its abscissa. Therefore, the required distance is 5 units.
The co – ordinates of the mid – point of the line segment joining the points (– 2, 3) and (4, – 5) are
Let the coordinates of midpoint C(x,y) of the line segment joining the points A(−2,3) and B(4,−5).
The point where the perpendicular bisector of the line segment joining the points A(2, 5) and B(4, 7) cuts is:
Since, the point, where the perpendicular bisector of a line segment cuts, is the midpoint of that line segment.
∴ Coordinates of Midpoint of line segment AB =
The distance of a point from the y – axis is called
The distance of a point from the y – axis is called abscissa.
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