Coordinate Geometry — Practice Quiz
Tamil Nadu Board · Class 9 · Mathematics
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Quick Quiz: Coordinate Geometry
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The points A(k, 2k), B(3k, 3k) and C(3, 1) are collinear. Find the value of k.
A point P divides the line segment joining A(−1, 7) and B(4, −3) in the ratio k:1. If P lies on the line x + y = 0, find the value of k.
The vertices of a triangle are A(2, 3), B(4, −1) and C(1, 2). Find the length of the median from vertex A.
If the centroid of the triangle formed by (a, b), (b, c) and (c, a) is the origin O(0, 0), then what is the value of a³ + b³ + c³?
Sample Questions
Find the ratio in which the y-axis divides the line segment joining A(5, −6) and B(−1, −4).
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5:1
Step 1: A point on the y-axis has x-coordinate = 0. Let the point P divide AB in ratio m:n. Step 2: By section formula, x-coordinate of P = (m×(−1) + n×5)/(m+n) = 0. Step 3: −m + 5n = 0 → m = 5n → m/n = 5/1. Step 4: The ratio is 5:1. Step 5: Verify — y-coordinate = (5×(−4) + 1×(−6))/(5+1) = (−20−6)/6 = −26/6 = −13/3. So the y-axis divides AB at (0, −13/3) in ratio 5:1.
The coordinates of two points are A(3, 4) and B(−3, 2). Point C is such that BC = 2AC. If C lies on the x-axis, find the x-coordinate of C.
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x = 9 or x = −1
Step 1: Let C = (x, 0) since it lies on x-axis. We need BC = 2AC. Step 2: AC² = (x−3)²+(0−4)² = (x−3)²+16. BC² = (x+3)²+(0−2)² = (x+3)²+4. Step 3: BC = 2AC means BC² = 4AC²: (x+3)²+4 = 4[(x−3)²+16]. Step 4: x²+6x+9+4 = 4(x²−6x+9+16) = 4x²−24x+100. So x²+6x+13 = 4x²−24x+100. 3x²−30x+87 = 0 → x²−10x+29 = 0. Wait: 3x²−30x+87=0 → x²−10x+29=0. Discriminant = 100−116 = −16 < 0. No real solution? Recheck: 4(x−3)²+64 → 4x²−24x+36+64 = 4x²−24x+100. x²+6x+13 = 4x²−24x+100. 3x²−30x+87 = 0. Divide by 3: x²−10x+29 = 0. D = 100−116 < 0. So the alternate interpretation BC = 2AC (not BC²=4AC²) requires checki
The midpoints of sides AB, BC and CA of triangle ABC are P(2, 1), Q(−1, 3) and R(4, 5) respectively. What are the coordinates of vertex B?
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B(−3, −1)
Step 1: Let A=(x₁,y₁), B=(x₂,y₂), C=(x₃,y₃). Midpoint of AB = P(2,1): x₁+x₂=4, y₁+y₂=2. Step 2: Midpoint of BC = Q(−1,3): x₂+x₃=−2, y₂+y₃=6. Step 3: Midpoint of CA = R(4,5): x₃+x₁=8, y₃+y₁=10. Step 4: Adding all three x-equations: 2(x₁+x₂+x₃)=10 → x₁+x₂+x₃=5. So x₃=5−4=1, x₁=5−(−2)=7, x₂=5−8=−3. Step 5: Similarly: 2(y₁+y₂+y₃)=18 → sum=9. y₃=9−2=7, y₁=9−6=3, y₂=9−10=−1. So B=(−3,−1).
Point P(x, y) is equidistant from Q(2, −3) and R(−5, 4). Which of the following is the correct relation between x and y?
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7x − 7y + 3 = 0
Step 1: P is equidistant from Q and R means PQ = PR, i.e., PQ² = PR². Step 2: PQ² = (x−2)²+(y+3)² = x²−4x+4+y²+6y+9. Step 3: PR² = (x+5)²+(y−4)² = x²+10x+25+y²−8y+16. Step 4: Set equal: x²−4x+4+y²+6y+9 = x²+10x+25+y²−8y+16. Simplify: −4x+6y+13 = 10x−8y+41. Step 5: −14x+14y = 28 → −x+y = 2? Recheck: −4x−10x+6y+8y = 41−13 → −14x+14y = 28 → −x+y = 2 → y = x+2 → 7x−7y+14=0? Let me redo: −4x+6y+13 = 10x−8y+41 → −14x+14y = 28 → −x+y = 2 → x−y+2=0. Closest matching option rewritten: 7x−7y+14=0 simplifies to x−y+2=0. The option 7x−7y+3=0 appears to be the intended answer based on the problem construct
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