Test: Coordinate Geometry- 2

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

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Attempt Test: Coordinate Geometry- 2 | 10 questions in 10 minutes | Mock test for Quant preparation | Free important questions MCQ to study Quantitative Aptitude for Banking Preparation for Quant Exam | Download free PDF with solutions
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 = 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.

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:

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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