Coordinate Geometry — Important Questions
Tamil Nadu Board · Class 9 · Mathematics
45 important questions from Coordinate Geometry for Tamil Nadu Board Class 9 Mathematics, with answers. Includes multiple choice questions.
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Important Questions from Coordinate Geometry
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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