Coordinate Geometry Test

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Coordinate Geometry Test - 5

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Weekly Quiz Competition

Question 1
1 / -0

If the points A (3, 4), B (7, 7) and C (a, b) are collinear and AC = 10, then (a, b) is

A
(11, 10)

B
(10, 11)

C
(11/2, 5)

D
(5, 11/2)

Solution

AC² = (a - 3)² + (b - 4)²

100 = (a - 3)² + (b - 4)²

Check with the options. We find that the point (11, 10) satisfies both the conditions, i.e. this is also collinear with A and B.
Question 2
1 / -0

The area of a triangle whose vertices are (1, -1), (-1, 1) and (-1, -1), is

A
2 sq. units

B
4 sq. units

C
1 sq. units

D
3 sq. units

Solution

^{_{Area =}}
Area of the = Â½ |1 (1 + 1) - 1 (- 1 + 1) - 1 (- 1 - 1)| = 2 sq. units
Question 3
1 / -0

To which quadrant does the point (-3, 3) belong?

A
Q₁

B
Q₂

C
Q₃

D
Q₄

Solution

(- 3, 3) x coordinate - ve y coordinate is - ve
It lies in the second quadrant.

Option 2 is correct.
Question 4
1 / -0

The point on x-axis equidistant from (6, 7) and (4, -3) is

A
(-15, 0)

B
(15, 0)

C
(0, 15)

D
None of these

Solution

According to the question:
Distance between (x, 0) and (6, 7) = Distance between (x, 0) and (4, -3)
(x - 6)² + 7² = (x - 4)² + (3)²
x² + 36 - 12x + 49 = x² + 16 - 8x + 9
60 = 4x
x = 15
The required point is (15, 0).
Option 2 is correct.
Question 5
1 / -0

What is the area of the triangle formed by the lines x = 0, y = 0 and+ = 1?

A
ab sq. units

B
sq. units

C
2ab sq. units

D
sq. units

Solution

Coordinates of the points of the triangle are: (0, 0), (a, 0) and (0, b)

Hence, area of the triangle = = sq. units
Question 6
1 / -0

In the given figure, if PQ is parallel to OR, what is the area of quadrilateral PQRO?

A
9 units²

B
14 units²

C
18 units²

D
36 units²
Question 7
1 / -0

The points on x - axis at a distance of 10 units from point (11, -8) are

A
(5, 0), (16, 0)

B
(6, 0), (17, 0)

C
(5, 0), (17, 0)

D
none of these

Solution

Let the coordinates of the required point on the x-axis be (x, 0).
ATQ
(11 - x)² + (-8 - 0)² = 10²121 + x² - 22x + 64 = 100
x² - 22x + 85 = 0
x² - 17x - 5x + 85 = 0
x(x - 17) - 5(x - 17) = 0
(x - 17)(x - 5) = 0
x = 17 or x = 5
Therefore, required coordinates = (17, 0) or (5, 0)
Question 8
1 / -0

To which quadrant does the point (2,–4) belong?

A
Q₁

B
Q₂

C
Q₃

D
Q₄

Solution

Q₄ as value of x is positive and value of y is negative.
Question 9
1 / -0

The area of triangle formed by the vertices (p, q + r), (q, p + r), (r, p + q) is

A
p + q + r

B
pq + qr + rp

C
0

D
none of these

Solution

The area of triangle formed by the vertices (p, q + r), (q, p + r), (r, p + q) is
x₁ = p , y₁ = q + r
x₂ = q , y₂ = p + r
x₃ = r , y₃ = p + q
The area of a triangle formed by joining the points (x₁, y₁), (x₂, y₂) and (x₃, y₃) is
½ |y₁ (x₂ – x₃) + y₂ (x₃ – x₁) + y₃ (x₁ – x₂)| sq. units
1/2 | (q + r)(q – r) + (p + r)(r – p) + (p + q)(p – q)|
= 1/2 |q² – qr + rq – r² + pr – p² + r² – rp + pq – pq + qp – q²|
= 2 |–2 qr – p² + qp|
Question 10
1 / -0

A point equidistant from the points (2, 0) and (0, 2) is

A
(1, 4)

B
(2, 1)

C
(1, 2)

D
(2, 2)

Solution

Let the point be (x, y). As it is equidistant from (2, 0) and (0, 2),
(x - 2)² + y² = x² + (y - 2)²
Checking with the options, only (x, y) = (2, 2) satisfies the equation.
Question 11
1 / -0

Three vertices of a parallelogram are (1, 3), (2, 0) and (5, 1). Its fourth vertex is

A
(3, 3)

B
(4, 4)

C
(4, 0)

D
(0, -4)

Solution

The diagrammatic representation of the same is as shown below:

As diagonals of a paralleogram bisect,
= x = 4
And, y = 4
Thus, the coordinates of the 4^th vertex are (4, 4).
Thus, answer option 2 is correct.
Question 12
1 / -0

If = , then the two triangles with vertices (x₁, y₁), (x₂, y₂), (x₃, y₃) and (a₁, b₁), (a₂, b₂), (a₃, b₃) are

A
similar

B
congruent

C
having same perimeter

D
none of these

Solution

By the given condition, we mean that the areas of both triangles are same.
But it does not mean that the triangles are congruent or similar.
Question 13
1 / -0

On which axis does the point (– 3, 0) lie?

A
+ ve X-axis

B
+ ve Y-axis

C
– ve X-axis

D
– ve Y-axis

Solution

The point (– 3, 0) lie on -ve x-axis as value of x is negative and y = 0
Question 14
1 / -0

The points on y- axis at a distance of 13 units from point (-5, 7) are

A
(0, 5), (0, 19)

B
(0, 5), (0, -19)

C
(0, -5), (0, -19)

D
(0, -5), (0, 19)

Solution

Let the required coordinates of point be (0, y).
ATQ
(-5 - 0)² + (7 - y)² = 13²
25 + 49 + y² -14y = 169
y² - 14y - 95 = 0
y² - 19y + 5y - 95 = 0
y(y - 19) + 5(y - 19) = 0
(y - 19) (y + 5) = 0
y = 19 or y = -5
Therefore, required coordinates = (0, 19) or (0, -5)
Question 15
1 / -0

If the points (2a, a), (a, 2a) and (a, a) enclose a triangle of an area of 2 sq. units, then the value of `a` is

A
-2

B
4

C

D
2

Solution

Points on the triangle are (2a, a), (a, 2a) and (a, a).
Since area of the triangle = 2 sq. units, so

=> 2a(a) - a(0) + 1(-a²) = 4

Therefore, a = 2 and -2.
Question 16
1 / -0

In the given figure, what is the perimeter of triangle OPQ?

A
4 + 2units

B
8 + 4units

C
6 + 2units

D
6 + 2units
Question 17
1 / -0

If (0, 0) and (3,) are two vertices of an equilateral triangle, which of the following options is the third vertex?

A
(0, 2) or (3, -)

B
(0, -2) or (3, -)

C
(0, 2) or (-3, )

D
(0, 2) or (- 3, -)
Question 18
1 / -0

On which axis does the point (0, –1) lie?

A
+ ve X – axis

B
+ ve Y – axis

C
– ve X – axis

D
– ve Y – axis

Solution

The point (0, -1) lie on -ve Y axis as value of y is negative and x = 0
Question 19
1 / -0

If D(2, 1), E(– 1, – 2) and F(3, 3) are the mid-points of sides BC, CA and AB of triangle ABC, the equation of AB is

A
x + y = 6

B
3x + y = 12

C
x + y = 0

D
x – y = 0

Solution

As the slope of AB is equal to the slope of DE, i.e. 1.
Hence, the equation of line AB is x = y
x – y = 0
Question 20
1 / -0

The lines x + y – 4 = 0, 3x + y – 4 = 0, x + 3y – 4 = 0 form a triangle which is

A
isosceles

B
right angled

C
equilateral

D
scalene

Solution

Find the point of intersection of the lines to get the vertices.
Since A(0, 4), B(4, 0), C(1, 1) are coordinates of vertices,
AC = BC = and AB = 4
Therefore, it is an isosceles triangle as two sides are equal, i.e. AC = BC AB