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If the distance between the points (2,  2) and (1, x) is 5, one of the values of x is
(5)²= [{2(1)}² + {x(2)}²]
25= 9+(x+2)²
259 = (x+2)²
√16=x+2
x+2= 4
x=42
x= 2.
A triangle with vertices (4, 0), ( 1,  1) and (3, 5) is a/an
Let A(4, 0), B(1, 1), C(3, 5)
⇒ AB^{2} + AC^{2} = BC^{2} and AB = AC
Hence, triangle is an isosceles rightangled triangle.
The points A (9, 0), B (9, 6), C (9, 6) and D (9, 0) are the vertices of a
AB= √(99)²+(60)²=6
BC= √(99)²+(66)²=18
CD = √(9+9)²+(06)²=6
DA = √(99)²+(00)²=18
so,it's opposite sides are equal,hence it is rectangle
The points ( 4,0), (4, 0) and (0,3) are the vertices of a/an
Isosceles triangle.
The distance between the points A (0, 6) and B (0, 2) is
Since both these points lie on a straight line i.e x axis, distance will be the difference between the respective y coordinates
(0,2) & (0,6) => 6(2)) = 6+2 = 8
A circle drawn with origin as the centre passes through The point which does not lie in the interior of the circle is
Distance of from centre of the circle i. e.
The distance between the points A (0, 7) and B (0, 3) is
Since both these points lie on a straight line i.e x axis, distance will be the difference between the respective y coordinates
(0,3) (0,7)
7(3) = 7+3 = 10
If the distance between the points (4,p) and (1,0) is 5 units, then the value of p is
⇒ 3^{2} + p^{2} = 5^{2} ⇒ p^{2} = 25  9 = 16 ⇒ p = ± 4
The end points of diameter of circle are (2, 4) and (3, 1). The radius of the circle is
radius = Diameter/2
diameter = √((2(3))^{2 }+ (4(1))^{2})
= √5^{2}+5^{2} = √50 = 5√2
If P(1, 2), Q(4,6), R(5,7) and S(a, b) are the vertices of a parallelogram PQRS, then
Diagonals of parallelogram bisect each other.
456 tests

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