Class 9 Maths - Coordinate Geometry Previous Year Questions

Sol:

The line ∠M can be above or below the x-axis

∴ coordinates of P are (3, 5) or (–3, 5)

Sol: Only the statement (i) is true.

Sol: The y-coordinate of a point in III-quadrant is always negative.

Sol: ∵ abscissa of A = – 2 and abscissa of B = – 3
∵ (– 2) – (– 3) = –2 + 3 =1

Sol: ∵ The coordinates have opposite signs in II and IV quadrants.
∴ the given point lies in I or III quadrant.

Sol: The line ∠M can be above or below the x-axis
∴ coordinates of P are (3, 5) or (–3, 5)

Sol: (i) The abscissa of P is –3.
(ii) The ordinate of Q is –2.
(iii) The ordinate of R is 2.
(iv) The abscissa of S is –5.

Sol: (i) The point B is having its ordinate as (–5).
(ii) The point C is having its abscissa as 4.
(iii) The coordinates of O are (0, 0).
(iv) The point A is having abscissa as (–3).

Sol: 

(i) ∵ The abscissa of the point Q is 6.  
The abscissa of the point R is 6.
∴The points Q and R have the same abscissa.
(ii) ∵ The ordinate of P is (–3).  
The ordinate of R is (–3).
∴ The points P and R have the same ordinate.

Sol:

The vertices of the triangle are A, B and C.
The coordinates of A are (–6, 4).
The coordinates of B are (–3, –3).
The coordinates of C are (2, 2).

Sol:

The vertices of the quadrilateral are P, Q, R and S.
The coordinates of P are (–4, 4).
The coordinates of Q are (8, 2).
The coordinates of R are (6, –3).
The coordinates of S are (–2, –2).

Sol:

Let A (3,0), B (6, 0), C (x, y).

∴ ∆ABC is an equilateral

∴ AB = BC = AC

AB2 = BC2 = AC2 …[Squaring throughout

(6 – 3)² + (0 – 0)² = (x – 6)² + (1 – 0)² = (x – 3)² + (y – 0)²

9 = x² – 12x + 36 + y² = x² – 6x + 9 + y²

Sol:

Area of ∆ = 15 sq. units

12 [1(p – 7) + 4(7 + 3)) + (-9)(-3 – p)] = ±15

p – 7 + 40 + 27 + 9p = ±30

10p + 60 = ± 30

10p = -60 ± 30

p = −60±30 / 10

∴ Taking +ve sign, p = −60 + 30 / 10 = −30 / 10 = -3

Taking -ve sign, p = −60 − 30 / 10 = −90 / 10 = -9

Sol:

Area of ∆ABD

= 1/2 (x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]

= 1/2[4(6 – 5) + 7(5 – 2) + 4(2 – 6))

= 1/2(4 + 21 – 16) = 9/2 sq.units …(i)

Area of ∆ADC

= 1/2 [4(5 – 4) + 4(4 – 2) + 162 – 5)]

= 1/2(4 + 8 – 3) = 9/2sq.units

From (i) and (ii),

Area of ∆ABD = Area of ∆ADC

∴ Median AD divides ∆ABC into two triangles of equal area.

Sol:
Let A(x, y), B(1, 2), C(2, 1).
Area of ∆ABC = 6 sq. units …[Given
As 1/2 [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)] = 6
∴ x(2 – 1) + 1(1 – y) + 2(y – 2) = ±12
x + 1 – y + 2y – 4 = ±12
Taking +ve sign
x + y = 12 + 4 – 1
∴ x + y = 15
Taking -ve sign
x + y = -12 + 4 – 1
∴ x + y = -9

Sol:

Sol:

Sol:
PQ = PR …[GivenPQ2 = PR2 … [Squaring both sides∴ (7 – 2)² + (0 – 4)² = (x – 2)² + (9 – 4)²
⇒ 25 + 16 = (x – 2)² + 25
⇒ 16 = (x – 2)²
⇒ ±4 = x – 2 …[Taking sq. root of both sides
⇒ 2 ± 4 = x
⇒ x = 2 + 4 = 6 or x = 2 – 4 = -2

Sol:
Centroid = 

a + b + c = 0
If a + b + c = 0
then, as we know
a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ac)
∴ a³ + b³ + c³ – 3abc = 0 … [Since a + b + c = 0
∴ a³ + b³ + c³ = 3abc …(Hence proved)

Sol:
Let A(k + 1, 1), B(4, -3) and C(7, -k).
We have, Area of ∆ABC = 6 … [Given
6 = 1/2 [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
6 = 1/2[(k + 1)(-3 + k) + 4(-k – 1) + 7(1 + 3)]
12 = (-3k + k2 – 3 + k – 4k – 4 + 28]
12 = [k2 – 6k + 21]
⇒ k2 – 6k + 21 – 12 = 0
⇒ k2 – 6k + 9 = 0
⇒ k2 – 3k – 3k + 9 = 0
⇒ k(k – 3) – 3(k – 3) = 0 =
⇒ (k – 3) (k – 3) = 0
⇒ k – 3 = 0 or k – 3 = 0
⇒ k = 3 or k = 3
Solving to get k = 3.

Sol:
Let A(4, -6), B(3, -2) and C(5, 2) be the vertices of ∆ABC.
Since AD is the median
∴ D is the mid-point of BC.
⇒ D(3 + 5 / 2,−2 + 2 / 2) ⇒ D(4,0)
Area of ∆ABD
= 1/2 [4(-2 – 0) + 3(0 + 6) + 4(-6 + 2)]
= 1/2 [-8 + 18 – 16) = 1/2 [-6] = -3
But area of A cannot be negative.
∴ ar(∆ABD) = 3 sq.units …(i)
Area of ∆ADC
= 1/2 [4(0 – 2) + 4(2 + 6) + 5(-6 – 0)]
= 1/2(-8 + 32 – 30] = 12 [-6] = -3
But area of ∆ cannot be negative.
∴ ar(∆ADC) = 3 sq.units
From (i) and (ii),
∴ Median AD of AABC divides it into two ∆s of equal area.