Test: Coordinate Geometry- 2


10 Questions MCQ Test Quantitative Aptitude for Banking Preparation | Test: Coordinate Geometry- 2


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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. 

Solution:

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

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). 

 

Solution:

AB = √(2+4)2 + (-5-7)2

= √62 + (-12)2

= √36+144 = √180

=√36*5 = 6√5 units.

QUESTION: 3

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

Solution:

AB = √(b-0)2-(0-a)2 

= √b2+a2 

= √a2+b2.

QUESTION: 4

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

Solution:

√(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

QUESTION: 5

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

Solution:

AB2 = (2 - 5)2 + (3 + 7)2 

=> (-3)2 + (10)2 

=> 9 + 100 => √109

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). 

Solution:

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 

QUESTION: 7

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

Solution:

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

QUESTION: 8

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

Solution:

AB2= (-5 - 0)2 + (-3 - 0)2 = 16 + 9 = 25 

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.

QUESTION: 9

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

Solution:

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

QUESTION: 10

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

Solution:

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