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Test: Coordinate Geometry- 2


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Test: Coordinate Geometry- 2 - Question 1

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 = √x2 + y2

Find the distance of the point A(4, -2) from the origin. 

Detailed Solution for Test: Coordinate Geometry- 2 - Question 1

OA = √4 - 02+(-2 - 0)2 = √16+4 = √20 = radic;4*5 = 2√5 units

Test: Coordinate Geometry- 2 - Question 2

Distance between two points :

If (x1, y1) and B(x2, y2) be two points, then AB = √(x2 - x1)2 + (y2 - y1)2

Find the distance between the points A(-4, 7) and B(2, -5). 

 

Detailed Solution for Test: Coordinate Geometry- 2 - Question 2

AB = √(2+4)2 + (-5-7)2

= √62 + (-12)2

= √36+144 = √180

=√36*5 = 6√5 units.

Test: Coordinate Geometry- 2 - Question 3

The distance between the points A(b, 0) and B(0, a) is. 

Detailed Solution for Test: Coordinate Geometry- 2 - Question 3

AB = √(b-0)2-(0-a)2 

= √b2+a2 

= √a2+b2.

Test: Coordinate Geometry- 2 - Question 4

If the distance of the point P(x, y) from A(a, 0) is a + x, then y2 = ? 

Detailed Solution for Test: Coordinate Geometry- 2 - Question 4

√(x-a)2+(y-0)2 = a + x 

= (x-a)2+y2 

= (a+x)2 => y2 = (x-a)2-(x-a)2-4ax => y2 = 4ax

Test: Coordinate Geometry- 2 - Question 5

The distance between the points A(5, -7) and B(2, 3) is:

Detailed Solution for Test: Coordinate Geometry- 2 - Question 5

AB2 = (2 - 5)2 + (3 + 7)2 

=> (-3)2 + (10)2 

=> 9 + 100 => √109

Test: Coordinate Geometry- 2 - Question 6

Area of a triangle : 

If A(x1,y1), B(x2,y2 and C(x3, y3) be three vertices of a ΔABC, then its area is given by:

Δ = 1/2 [x1(y2 - y3 + x2(y3 - y1) + x3(y1 - y2)]

Find the area of ΔABC whose vertices are A(9, -5), B(3, 7) and (-2, 4). 

Detailed Solution for Test: Coordinate Geometry- 2 - Question 6

Here, x1 = 9, x2 = 3, x3 = -2 and y1 = -5, y2 = 7, y3 = 4

= 1/2 [9(7-4) + 3(4+5) + (-2)(-5-7)] 

= 1/2 [9(3) + 3(9) - 2(-12)] 

= 1/2 [27 + 27 + 24] 

= 1/2 [78] 

= 39 sq.units 

Test: Coordinate Geometry- 2 - Question 7

Find the area of ΔABC whose vertices are A(2, -5), B(4, 9) and (6, -1). 

Detailed Solution for Test: Coordinate Geometry- 2 - Question 7

Here, x1 = 2, x2 = 4, x3 = 6 and y1 = -5, y2 = 9, y3 = -1

= 1/2 [2(9+1) + 4(-1+5) + 6(-5-9)] 

= 1/2 [2(10) + 4(4) + 6(-14)] 

= 1/2 [20 + 16 - 84] 

= 1/2 [-48] 

= 24 sq.units

Test: Coordinate Geometry- 2 - Question 8

The points A(0, 6), B(-5, 3) and C(3, 1) are the vertices of a triangle which is ? 

Detailed Solution for Test: Coordinate Geometry- 2 - Question 8

AB2= (-5 - 0)2 + (-3 - 0)2 = 25 + 9 = 34

BC2 = (3 + 5)2 + (1-3)2 = 82 + (-2)2 = 64 + 4 = 68 

AC2 = (3 - 0)2 + (1 - 6)2 = 9 + 25 = 34. 

AB = AC. ==> ΔABC is isosceles.

Test: Coordinate Geometry- 2 - Question 9

The points A(-4, 0), B(1, -4), and C(5, 1) are the vertices of 

Detailed Solution for Test: Coordinate Geometry- 2 - Question 9

AB2 = (1 + 4)2 + (-4 - 0)2 

= 25 + 16 = 41, 

BC2 = (5 - 1)2 + (1 + 4)2 = 42 + 52

= 16 + 25 = 41 

AC2 = (5 + 4)2 + (1 - 0)2 

= 81 + 1 = 82 

AB = BC and AB2 = BC2 = AC2 

ΔABC is an isosceles right angled triangle

Test: Coordinate Geometry- 2 - Question 10

 Find the coordinates of the point equidistant from the points A(1, 2), B (3, –4) and C(5, –6).

Detailed Solution for Test: Coordinate Geometry- 2 - Question 10

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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