Distance of a point from the origin:
The distance of a point A(x, y) from the origin O(0, 0) is given by OA = √x^{2} + y^{2}
Find the distance of the point A(4, 2) from the origin.
OA = √4  0^{2}+(2  0)^{2} = √16+4 = √20 = radic;4*5 = 2√5 units
Distance between two points :
If (x_{1}, y_{1}) and B(x_{2}, y_{2}) be two points, then AB = √(x_{2}  x_{1})^{2} + (y_{2}  y_{1})^{2}
Find the distance between the points A(4, 7) and B(2, 5).
AB = √(2+4)^{2} + (57)^{2}
= √6^{2} + (12)^{2}
= √36+144 = √180
=√36*5 = 6√5 units.
The distance between the points A(b, 0) and B(0, a) is.
AB = √(b0)^{2}(0a)^{2}
= √b^{2}+a^{2}
= √a^{2}+b^{2}.
If the distance of the point P(x, y) from A(a, 0) is a + x, then y^{2} = ?
√(xa)^{2}+(y0)^{2} = a + x
= (xa)^{2}+y^{2}
= (a+x)^{2} => y^{2} = (xa)^{2}(xa)^{2}4ax => y^{2} = 4ax
The distance between the points A(5, 7) and B(2, 3) is:
AB^{2} = (2  5)^{2} + (3 + 7)^{2}
=> (3)^{2} + (10)^{2}
=> 9 + 100 => √109
Area of a triangle :
If A(x_{1},y_{1}), B(x_{2},y_{2} and C(x_{3}, y_{3}) be three vertices of a ΔABC, then its area is given by:
Δ = 1/2 [x_{1}(y_{2}  y_{3} + x_{2}(y_{3}  y_{1}) + x_{3}(y_{1}  y_{2})]
Find the area of ΔABC whose vertices are A(9, 5), B(3, 7) and (2, 4).
Here, x_{1} = 9, x_{2} = 3, x_{3} = 2 and y_{1} = 5, y_{2} = 7, y_{3} = 4
= 1/2 [9(74) + 3(4+5) + (2)(57)]
= 1/2 [9(3) + 3(9)  2(12)]
= 1/2 [27 + 27 + 24]
= 1/2 [78]
= 39 sq.units
Find the area of ΔABC whose vertices are A(2, 5), B(4, 9) and (6, 1).
Here, x_{1} = 2, x_{2} = 4, x_{3} = 6 and y_{1} = 5, y_{2} = 9, y_{3} = 1
= 1/2 [2(9+1) + 4(1+5) + 6(59)]
= 1/2 [2(10) + 4(4) + 6(14)]
= 1/2 [20 + 16  84]
= 1/2 [48]
= 24 sq.units
The points A(0, 6), B(5, 3) and C(3, 1) are the vertices of a triangle which is ?
AB^{2}= (5  0)^{2} + (3  0)^{2} = 25 + 9 = 34
BC^{2} = (3 + 5)^{2} + (13)^{2} = 8^{2} + (2)^{2} = 64 + 4 = 68
AC^{2} = (3  0)^{2} + (1  6)^{2} = 9 + 25 = 34.
AB = AC. ==> ΔABC is isosceles.
The points A(4, 0), B(1, 4), and C(5, 1) are the vertices of
AB^{2} = (1 + 4)^{2} + (4  0)^{2}
= 25 + 16 = 41,
BC^{2} = (5  1)^{2} + (1 + 4)^{2} = 4^{2} + 5^{2}
= 16 + 25 = 41
AC^{2} = (5 + 4)^{2} + (1  0)^{2}
= 81 + 1 = 82
AB = BC and AB^{2} = BC^{2} = AC^{2}
ΔABC is an isosceles right angled triangle
Find the coordinates of the point equidistant from the points A(1, 2), B (3, –4) and C(5, –6).
Given three points A(1,2) B(3,4) and C(5,6)
we have to find the coordinates of the point equidistant from the points.
The point that is equidistant from three points is called circumcenter which can be evaluated to find the perpendicular bisectors.
To find the perpendicular bisectors of AB:
(11,2)
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