Coordinate Geometry Sample Paper - Class 10 Maths

Time: 1 hour

M.M. 30

Attempt all questions.

• Question numbers 1 to 5 carry 1 mark each.
• Question numbers 6 to 8 carry 2 marks each.
• Question numbers 9 to 11 carry 3 marks each.
• Question number 12 & 13 carry 5 marks each.

Q1: If the distance between the points A(2, -2) and B(-1, x) is equal to 5, then the value of x is: (1 Mark)
(a) 2
(b) -2
(c) 1
(d) -1

Ans: (a)

Sol:

Distance AB = √[(-1 - 2)² + (x - (-2))²] = √[(-3)² + (x + 2)²] = √[9 + (x + 2)²].

Given √[9 + (x + 2)²] = 5 ⇒ 9 + (x + 2)² = 25.

(x + 2)² = 16 ⇒ x + 2 = ±4 ⇒ x = 2 or x = -6.

Only x = 2 is among the options. Hence x = 2.

By distance formula, we know:

Take square root on both the sides,

2 + x = 4

x = 2
Q2: The midpoint of a line segment joining two points A(2, 4) and B(-2, -4) is (1 Mark)
(a) (-2, 4)
(b) (2, -4)
(c) (0, 0)
(d) (-2, -4)

Ans: (c)

Sol:

Midpoint M of A(2, 4) and B(-2, -4) is given by M = ((x₁ + x₂)/2, (y₁ + y₂)/2).

x-coordinate = (2 + (-2))/2 = 0/2 = 0.

y-coordinate = (4 + (-4))/2 = 0/2 = 0.

Hence the midpoint is (0, 0).

As per midpoint formula, we know;

x-coordinate of the midpoint = [2 + (-2)]/2 = 0/2 = 0

y-coordinate of the midpoint = [4 + (-4)]/2=0/2=0

Hence, (0, 0) is the midpoint of AB.
Q3: The distance of point A(2, 4) from the x-axis is _______________________________. (1 Mark)

Ans: 4 units

The distance of a point from the x-axis equals the absolute value of its y-coordinate. For A(2, 4), this distance is |4| = 4 units.

The distance of a point from the x-axis is equal to the ordinate of the point.
Q4: If O(p/3, 4) is the midpoint of the line segment joining the points P(-6, 5) and Q(-2, 3), the  the value of p is: (1 Mark)
(a) 7/2
(b) -12
(c) 4
(d) -4

Ans: (b)

Sol:

Midpoint of P(-6, 5) and Q(-2, 3) is ((-6 + (-2))/2, (5 + 3)/2) = (-8/2, 8/2) = (-4, 4).

Given midpoint is (p/3, 4). Thus p/3 = -4 ⇒ p = -12.

Therefore, the value of p is -12.

Since, (p/3, 4) is the midpoint of line segment PQ, thus;

p/3 = (-6-2)/2

p/3 = -8/2

p/3 = -4

p= -12

Therefore, the value of p is -12.
Q5: The ratio in which the line segment joining the points P(-3, 10) and Q(6, -8) is divided by O(-1, 6) is: (1 Mark)
(a) 1:3
(b) 3:4
(c) 2:7
(d) 2:5

Ans: (c)

Sol:

Let the ratio be k : 1, where O divides P(-3, 10) and Q(6, -8).

Using section formula for x-coordinate: x = (k·x₂ + x₁)/(k + 1) = (6k + (-3))/(k + 1).

Given x = -1, so (6k - 3)/(k + 1) = -1.

6k - 3 = -k - 1 ⇒ 7k = 2 ⇒ k = 2/7.

Thus the ratio is 2 : 7 (internally from P to Q).

Let k :1 be the ratio in which the line segment joining P( -3, 10) and Q(6, -8) is divided by point O(-1, 6).

By the section formula, we have;

-1 = ( 6k - 3)/(k + 1)

-k - 1 = 6k - 3

7k = 2

k = 2/7

Hence, the required ratio is 2:7.
Q6: Find a relation between x and y such that the point (x, y) is equidistant from the points (7, 1) and (3, 5). (2 Marks)

Ans: Let P(x, y) be equidistant from the points A(7, 1) and B(3, 5).

Then AP = BP ⇒ AP² = BP².

Using distance formula, (x - 7)² + (y - 1)² = (x - 3)² + (y - 5)².

Expand and cancel x² and y² terms:

x² - 14x + 49 + y² - 2y + 1 = x² - 6x + 9 + y² - 10y + 25.

Bring like terms together: -14x + 50 - 2y = -6x + 34 - 10y.

Rearrange: -8x + 16 + 8y = 0 ⇒ divide by 8 ⇒ -x + 2 + y = 0.

Therefore x - y = 2.

Hence, the relation between x and y is x - y = 2.

Q7: The point A(3, y) is equidistant from the points P(6, 5) and Q(0, -3). Find the value of y. (2 Marks)

Ans:

PA = QA (given) ⇒ PA² = QA².

(3 - 6)² + (y - 5)² = (3 - 0)² + (y + 3)².

9 + (y - 5)² = 9 + (y + 3)² ⇒ (y - 5)² = (y + 3)².

Take square roots: y - 5 = ±(y + 3).

Case 1: y - 5 = y + 3 ⇒ -5 = 3, impossible.

Case 2: y - 5 = -y - 3 ⇒ 2y = 2 ⇒ y = 1.

Therefore y = 1.

Q8: Name the type of triangle formed by the points A (-5, 6), B (-4, -2) and C (7, 5). (2 Marks)

Ans: The points are A (-5, 6), B (-4, -2) and C (7, 5).

Compute side lengths using distance formula:

AB = √[(-4 - (-5))² + (-2 - 6)²] = √[(1)² + (-8)²] = √[1 + 64] = √65.

BC = √[(7 - (-4))² + (5 - (-2))²] = √[(11)² + (7)²] = √[121 + 49] = √170.

CA = √[(7 - (-5))² + (5 - 6)²] = √[(12)² + (-1)²] = √[144 + 1] = √145.

All three sides have different lengths (√65, √170, √145). Hence triangle ABC is a scalene triangle.

Using distance formula,

Q9: Three vertices of a parallelogram taken in order are (-1, 0), (3, 1) and (2, 2) respectively. Find the coordinates of fourth vertex. (3 Marks)

Ans: Let A(-1, 0), B(3, 1), C(2, 2) and D(x, y) be the vertices of parallelogram ABCD taken in order. The diagonals of a parallelogram bisect each other, so the mid-point of AC equals the mid-point of BD.

Mid-point of AC = ((-1 + 2)/2, (0 + 2)/2) = (1/2, 1).

Mid-point of BD = ((3 + x)/2, (1 + y)/2). Set equal to mid-point of AC:

(3 + x)/2 = 1/2 ⇒ 3 + x = 1 ⇒ x = -2.

(1 + y)/2 = 1 ⇒ 1 + y = 2 ⇒ y = 1.

Therefore D(-2, 1).

∴ Coordinates of the fourth vertex, D(-2, 1).

Q10: If the point P(k - 1, 2) is equidistant from the points A(3, k) and B(k, 5), find the values of k. (3 Marks)

Ans:

Given PA = PB ⇒ PA² = PB².

PA² = (k - 1 - 3)² + (2 - k)² = (k - 4)² + (2 - k)².

PB² = (k - 1 - k)² + (2 - 5)² = (-1)² + (-3)² = 1 + 9 = 10.

So (k - 4)² + (2 - k)² = 10.

Expand: (k² - 8k + 16) + (k² - 4k + 4) = 10 ⇒ 2k² - 12k + 20 = 10.

⇒ 2k² - 12k + 10 = 0 ⇒ k² - 6k + 5 = 0.

Factor: (k - 5)(k - 1) = 0 ⇒ k = 5 or k = 1.

∴ k = 5 or k = 1.

Q11: Prove that the points A(0, -1), B(-2, 3), C(6, 7) and D(8, 3) are the vertices of a rectangle ABCD. (3 Marks)

Ans:

We show that opposite sides are equal and parallel (so ABCD is a parallelogram), and that the diagonals are equal (which implies a rectangle).

Compute vectors of sides:

AB = B - A = (-2 - 0, 3 - (-1)) = (-2, 4).

BC = C - B = (6 - (-2), 7 - 3) = (8, 4).

CD = D - C = (8 - 6, 3 - 7) = (2, -4).

DA = A - D = (0 - 8, -1 - 3) = (-8, -4).

Observe AB = (-2, 4) and CD = (2, -4) = -1·(-2, 4), so AB ∥ CD and |AB| = |CD|.

Also BC = (8, 4) and DA = (-8, -4) = -1·(8, 4), so BC ∥ DA and |BC| = |DA|.

Since both pairs of opposite sides are parallel and equal, ABCD is a parallelogram.

Compute diagonals:

AC = C - A = (6 - 0, 7 - (-1)) = (6, 8) ⇒ |AC| = √(6² + 8²) = √(36 + 64) = √100 = 10.

BD = D - B = (8 - (-2), 3 - 3) = (10, 0) ⇒ |BD| = √(10² + 0²) = √100 = 10.

Diagonals AC and BD are equal in length. In a parallelogram, equal diagonals imply it is a rectangle.

Therefore ABCD is a rectangle.

Diag. AC = BD = 10

∴ ABCD is a rectangle. ... [∵ Opp. sides are equal & diagonals are also equal]

Q12: Find the ratio in which the line x - 3y = 0 divides the line segment joining the points (-2, -5) and (6, 3). Find the coordinates of the point of intersection. (5 Marks)

Ans: Let A(-2, -5) and B(6, 3). Suppose the line divides AB in the ratio k : 1, internal to A and B. Then the point of division P has coordinates

P = ((k·x₂ + x₁)/(k + 1), (k·y₂ + y₁)/(k + 1)) = ((6k - 2)/(k + 1), (3k - 5)/(k + 1)).

P lies on x - 3y = 0, so (6k - 2)/(k + 1) - 3·(3k - 5)/(k + 1) = 0.

Multiply both sides by (k + 1): 6k - 2 - 9k + 15 = 0 ⇒ -3k + 13 = 0 ⇒ k = 13/3.

Thus the ratio is 13 : 3 (from A toward B).

Coordinates of P:

x = (6·(13/3) - 2)/((13/3) + 1) = (26 - 2)/((16/3)) = 24 · (3/16) = 72/16 = 9/2.

y = (3·(13/3) - 5)/((13/3) + 1) = (13 - 5)/(16/3) = 8 · (3/16) = 24/16 = 3/2.

Therefore P = (9/2, 3/2).

The line x - 3y = 0 divides the line segment AB in the ratio 13 : 3 and the point of intersection is (9/2, 3/2).

Q13: Find the ratio in which P(4, m) divides the line segment joining the points A(2, 3) and B(6, -3). Hence, find m. (5 Marks)

Ans: Let P(4, m) divide AB in the ratio k : 1 (internal), where A(2, 3) and B(6, -3).

Using section formula for x-coordinate: (k·6 + 2)/(k + 1) = 4 ⇒ 6k + 2 = 4k + 4 ⇒ 2k = 2 ⇒ k = 1.

Thus P divides AB in ratio 1 : 1 (mid-point).

Now use y-coordinate: m = (k·(-3) + 3)/(k + 1) = (-3k + 3)/(k + 1).

Substitute k = 1: m = (-3 + 3)/(1 + 1) = 0/2 = 0.

Therefore the ratio is 1 : 1 and m = 0.

Let P(4, m) divides the line segment joining the points A(2, 3) and B(6, -3) in the ratio k:1.

Here, P(4, m) = (x, y)

A(2, 3) = (x₁, y₁)

B(6, -3) = (x₂, y₂)

Using section formula,

By equating the x-coordinate,

(6k + 2)/(k + 1) = 4

6k + 2 = 4k + 4

6k - 4k = 4 - 2

2k = 2

k = 1

Thus, the point P divides the line segment joining A and B in the ratio 1 : 1.

Now by equating the y-coordinate,

(-3k + 3)/(k + 1) = m

Substituting k = 1,

[-3(1) + 3]/(1 + 1) = m

m = (3 - 3)/2

m = 0