HOTS Questions on Class 10 Maths Chapter 7 ‘Coordinate Geometry’ with Answers

Higher-Order Thinking Skills (HOTS) question set for Chapter 7 ‘Coordinate Geometry’ for Class 10 is designed to push students beyond standard procedures and build deep, transferable problem-solving abilities in analytic geometry. Aligned with CBSE and NCERT learning objectives, these questions focus on conceptual understanding and application of ideas such as plotting and interpreting points, distance and section formulas, area of triangles using coordinates, slopes and equations of lines, and solve real-world problems. Each question includes a clear answer and brief reasoning to support classroom discussion and self-study. Download the PDF to access the same HOTS questions in a printable format, making offline learning and exam revision easier.

Solved HOTS Questions on Coordinate Geometry

Question 1: Point A(0, 2) is equidistant from points B(3, p) and C(p, 5). Find all possible values of p.

Solution:

Given A(0, 2), B(3, p), C(p, 5). Since A is equidistant, AB = AC.

Using the Distance Formula:

AB² = (3−0)² + (p−2)² = 9 + p² − 4p + 4 = p² − 4p + 13

AC² = (p−0)² + (5−2)² = p² + 9

Setting AB² = AC²:

p² − 4p + 13 = p² + 9 

⇒ −4p = −4 

⇒ p = 1

Question 2: Points P, Q, R, and S divide the line segment joining A(1, 2) and B(6, 7) into five equal parts. Find the coordinates of P and R.

Solution: 

Five equal parts means ratios from A are 1:4, 2:3, 3:2, 4:1 for P, Q, R, S respectively.

For P (ratio 1:4 from A):

x_P = (1×6 + 4×1)/(1+4) = 10/5 = 2 

y_P = (1×7 + 4×2)/(1+4) = 15/5 = 3 

P = (2, 3)

For R (ratio 3:2 from A):

x_R = (3×6 + 2×1)/(3+2) = 20/5 = 4 

y_R = (3×7 + 2×2)/(3+2) = 25/5 = 5 

R = (4, 5)

P = (2, 3) and R = (4, 5)

Question 3: In what ratio does the point (−4, 6) divide the line segment joining A(−6, 10) and B(3, −8)?

Solution: 

Let the required ratio be k:1. The dividing point is (−4, 6).

Using x-coordinate in the section formula:

−4 = (k×3 + 1×(−6)) / (k+1) 

 −4(k+1) = 3k − 6 

 −4k − 4 = 3k − 6 

 −4k − 3k = −6 + 4 

 −7k = −2 

⇒ k = 2/7

So the ratio is k:1 = 2/7 : 1 = 2:7.

The point divides AB in the ratio 2:7 internally.

Question 4: The vertices of a triangle are (1, k), (4, −3), and (−9, 7). If its area is 15 sq. units, find the value(s) of k.

Solution: 

Let (x₁,y₁) = (1,k), (x₂,y₂) = (4,−3), (x₃,y₃) = (−9,7).

Apply the area formula:

Area = ½|x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)| 

= ½|1(−3−7) + 4(7−k) + (−9)(k−(−3))| 

= ½|1(−10) + 4(7−k) + (−9)(k+3)| 

= ½|−10 + 28 − 4k − 9k − 27| 

= ½|−9 − 13k|

Set equal to 15:

½|−9 − 13k| = 15 |−9 − 13k| = 30

Case 1: −9 − 13k = 30 ⇒ −13k = 39 ⇒ k = −3

Case 2: −9 − 13k = −30 ⇒ −13k = −21 ⇒ k = 21/13

k = −3 or k = 21/13

Question 5: Find the relation between x and y if points (2, 1), (x, y), and (7, 5) are collinear.

Solution: 

For collinear points, area of triangle = 0:

½|2(y−5) + x(5−1) + 7(1−y)| = 0 

⇒ |2y−10 + 4x + 7 − 7y| = 0 

⇒ |4x − 5y − 3| = 0

So: 4x − 5y − 3 = 0, which gives us 4x − 5y = 3

Relation: 4x − 5y = 3

Question 6: If the points A(6, 1), B(8, 2), C(9, 4), and D(p, 3) are the vertices of a parallelogram taken in order, find the value of p.

Solution: 

Diagonals of a parallelogram bisect each other. Diagonals here are AC and BD.

Midpoint of AC:

= ((6+9)/2, (1+4)/2) = (15/2, 5/2)

Midpoint of BD:

= ((8+p)/2, (2+3)/2) = ((8+p)/2, 5/2)

Equating midpoints:

x: (8+p)/2 = 15/2 

8+p = 15 

⇒ p = 7

Question 7: Prove that the points A(3, 0), B(6, 4), and C(−1, 3) are the vertices of a right-angled isosceles triangle.

Solution: 

Compute all three sides:

AB = √[(6−3)²+(4−0)²] = √[9+16] = √25 = 5 

BC = √[(−1−6)²+(3−4)²] = √[49+1] = √50 

CA = √[(3−(−1))²+(0−3)²] = √[16+9] = √25 = 5

AB = CA = 5, so the triangle is isosceles.

Check the Pythagorean theorem :

AB² + CA² = 25 + 25 = 50 

BC² = 50

Since AB² + CA² = BC², the right angle is at vertex A.

△ ABC is a right-angled isosceles triangle (right angle at A).

Question 8: Points P and Q trisect the line segment joining A(−2, 0) and B(0, 8) such that P is nearer to A. Find the area of triangle OPQ, where O is the origin.

Solution: 

P divides AB in ratio 1:2 (nearer to A):

P = ((1×0 + 2×(−2))/3, (1×8 + 2×0)/3) = ((0−4)/3, 8/3) = (−4/3, 8/3)

Q divides AB in ratio 2:1:

Q = ((2×0 + 1×(−2))/3, (2×8 + 1×0)/3) = (−2/3, 16/3)

Area of △OPQ with O(0,0), P(−4/3, 8/3), Q(−2/3, 16/3):

= ½|0(8/3 − 16/3) + (−4/3)(16/3 − 0) + (−2/3)(0 − 8/3)| 

= ½|0 + (−4/3)(16/3) + (−2/3)(−8/3)| 

= ½|−64/9 + 16/9| 

= ½|−48/9| 

= ½ × 48/9 = 24/9 = 8/3

Area of triangle OPQ = 8/3 sq. units

Formula Cheat Sheet

• Distance Formula

d = √[(x₂−x₁)² + (y₂−y₁)²]

• Midpoint

M = ((x₁+x₂)/2, (y₁+y₂)/2)

• Section (Internal)

x = (mx₂+nx₁)/(m+n)

y = (my₂+ny₁)/(m+n)

• Area of △ = ½|x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)|

• Collinearity

Area of △ = 0

• Parallelogram Test

Midpt(AC) = Midpt(BD)

Questions PDF with worked-out examples for Class 10 Chapter 7: Coordinate Geometry, perfect for last-minute CBSE exam revision.

Class 10 Chapter 7: Coordinate Geometry HOTS PDF

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