Assertion & Reason Test: Coordinate Geometry - Class 10 MCQ

Here, A → (0 , 5) and B → (– 3 , 1)

= 5 units

∴ Assertion is correct.

In case of reason: If A (2, 7) lies on perpendicular bisector of P (6, 5) and Q (0, –4), then

AP = AQ

∴ By using Distance Formula,

As, AP ≠ AQ

Therefore, A does not lie on the perpendicular bisector of PQ.

∴ Reason is incorrect.

Hence, assertion is correct but reason is incorrect.

and section formula (externally), we have

So, Reason is correct.

Here, x₁ = -5, y₁ = 11, x₂ = 4, y₂ = -7, m₁ = 7, m₂ = 2

So, Assertion is also correct

∴ Distance between A (-2, 0) and B (2, 0),

[∵ distance between the points (x₁, y₁) and (x₂, y₂)

Similarly, distance between B (2, 0) and C (0, 2),

In △ABC, distance between C (0, 2) and A (-2, 0),

Distance between F (0, 4) and D (–4, 0),

Distance between F (0, 4) and E (4, 0),

and distance between E (4, 0) and D (–4, 0),

Here, we see that sides of DABC and DFDE are proportional.

Therefore, by SSS similarity rule, both the triangles are similar.

∴ Assertion is correct.

In case of reason: Point Q (6, 8) will lie outside the circle if its distance from the centre of circle is greater than the radius of the circle.

Distance between centre O (0, 0) and P (5, 0),

As, point P lies at the circle, therefore, OP = Radius of circle.

Distance between centre O (0, 0) and Q (6, 8),

As OQ > OP, therefore, point Q (6, 8) lies outside of the circle.

∴ Reason is correct.

Hence, both assertion and reason are correct but reason is not the correct explanation for assertion.

(−k + 1, 2k), (k, 2 − 2k),(−4 −k, 6 − 2k)

Since, these points are collinear. Therefore the area of triangle formed by the triangle formed by the points will be zero.

Therefore,

(−k + 1)(2 − 2k − 6 + 2k) + k(6 − 2k − 2k) + (−4 − k)(2k − 2 + 2k) = 0

(−k + 1)(−4) + k(6 − 4k) + (−4 − k)(4k − 2) = 0

4k − 4 + 6k − 4k² − 16k + 8 − 4k² + 2k = 0

−2k² − k + 1 = 0

2k² + k − 1 = 0

2k² + 2k − k − 1 = 0

2k(k + 1) − 1(k + 1) = 0

(k + 1)(2k − 1) = 0

k = −1 or k = 1/2.​

So, Reason is correct.

Given, the points A (4,3) and B (x, 5) lie on a circle with center O(2,3).

Then OA = OB ⇒(OA)² = (OB)²

⇒ (4 – 2)² + (3 – 3)² = (x – 2)² + (5 – 3)²

⇒ (2)² +(0)² = (x – 2)² + (2)² ⇒ 4 = (x – 2)² + 4 ⇒(x – 2)² = 0

⇒ x – 2 = 0 ⇒ x = 2

So, Assertion is correct

where, (x₁, y₁) = (4, p)

(x₂, y₂) = (1, 0)

And, d = 5

Put the values, we have

5² = (1 − 4)² + (0 – p)²

25 = (–3)² + (–p)²

25 – 9 = p²

16 = p²

+4, –4 = p

∴ Assertion is incorrect.

In case of reason:

Let (x, y) be the point

Here, x₁ = 7, y₁ = –6, x₂ = 3, y₂ = 4, m = 1 and n = 2

So, the required point lies in IV^th quadrant.

∴ Reason is correct.

Hence, assertion is incorrect but reason is correct.

a + b + c = 0

So, Reason is correct.

Now, AB = 5 ⇒ AB² = 25

⇒ (x – 5)² + (-1 - 3)² = 25

⇒ (x – 5)² = 25 – 16 = 9

⇒ x – 5 = ±3

⇒ x = 5 ± 3

⇒ x = 2, 8

So, Assertion is also correct

Let(x₁, y₁, z₁) = (0, 0, 0)

(x, y ,z) = (2, −3, 3)

(x₂, y₂, z₂) = (−2, 3, −3)

By section formula:

x = m₁ + m₂ m₁ x 2 + m₂ x 1

⇒ 2 = k + 1 k.(−2) + 1.(0)

⇒ k = 4 − 1 on neglecting the negative sign

we get, therefore ratio is 1:4.