MATHEMATICS (GEOMETRY) | FYUG Even Semester Exam, 2025

Course No: MATIDC-151 | 2nd Semester Full Marks: 70 | Pass Marks: 28 | Time: 3 Hours

UNIT-I: Basic Coordinate Geometry

Question 1 (a)

1 Mark

Question: Write the formula for distance between two points (x1, y1) and (x2, y2).

Distance d = √[(x2 - x1)² + (y2 - y1)²]

Question 1 (b)

1 Mark

Question: Find the middle point of the line joining the points (1, 2) and (-1, -2).

Solution:

  • Mid-point formula: M = ((x1 + x2)/2, (y1 + y2)/2)
  • Calculation: M = ((1 - 1)/2, (2 - 2)/2) = (0/2, 0/2) = (0, 0)

Final Answer: (0, 0)

Question 3 (a)

2 Marks

Question: The line joining the points (-6, 8) and (8, 6) is trisected; find the coordinates of the points of trisection.

Solution:

Trisection points divide the line in ratios 1:2 and 2:1.

Point 1 (1:2):

  • x = [1(8) + 2(-6)] / (1+2) = (8 - 12)/3 = -4/3
  • y = [1(6) + 2(8)] / (1+2) = (6 + 16)/3 = 22/3

Point 2 (2:1):

  • x = [2(8) + 1(-6)] / (2+1) = (16 - 6)/3 = 10/3
  • y = [2(6) + 1(8)] / (2+1) = (12 + 8)/3 = 20/3

UNIT-II: Straight Lines

Question 4 (b)

1 Mark

Question: Write the equation of a straight line parallel to x-axis.

Equation: y = k (where k is a constant)

Question 5 (b)

4 Marks

Question: Find the equation to the straight line, which passes through (4, -5) and which is parallel to 3x + 4y + 5 = 0.

Solution:

  • The equation of a line parallel to 3x + 4y + 5 = 0 is of the form 3x + 4y + k = 0.
  • Since it passes through (4, -5): 3(4) + 4(-5) + k = 0
  • 12 - 20 + k = 0 => -8 + k = 0 => k = 8

Final Answer: 3x + 4y + 8 = 0

Question 6 (a)

4 Marks

Question: If p is the length of perpendicular from the origin to the line whose intercepts on the axes are a and b, then show that 1/p² = 1/a² + 1/b².

Solution:

The intercept form of the line is: x/a + y/b = 1 or x/a + y/b - 1 = 0.

Perpendicular distance 'p' from (0,0) is:

p = |(0/a + 0/b - 1)| / √[(1/a)² + (1/b)²]

p = 1 / √[1/a² + 1/b²]

Squaring both sides: p² = 1 / [1/a² + 1/b²]

Taking reciprocal: 1/p² = 1/a² + 1/b². Hence Proved.

UNIT-III: Pair of Straight Lines

Question 7 (c)

1 Mark

Question: Show that the pair of lines 6x² - 5xy - 6y² + 14x + 5y + 4 = 0 are perpendicular.

Solution:

For a pair of lines ax² + 2hxy + by² + 2gx + 2fy + c = 0 to be perpendicular:

a + b = 0

From the equation: a = 6 and b = -6.

a + b = 6 + (-6) = 0. Since the sum is zero, the lines are perpendicular.

UNIT-IV: The Circle

Question 10 (a)

1 Mark

Question: Write the condition that an equation of second degree may represent a circle.

Solution:

The equation ax² + 2hxy + by² + 2gx + 2fy + c = 0 represents a circle if:

  • Coefficient of x² = Coefficient of y² (a = b)
  • Coefficient of xy = 0 (h = 0)

Question 12 (b)

4 Marks

Question: Find the equation of the circle which has its centre at (1, -3) and touches the straight line 2x - y - 4 = 0.

Solution:

  • The radius is the perpendicular distance from the centre (1, -3) to the tangent 2x - y - 4 = 0.
  • Radius r = |2(1) - (-3) - 4| / √[2² + (-1)²]
  • r = |2 + 3 - 4| / √5 = 1 / √5
  • Equation of circle: (x - 1)² + (y + 3)² = (1/√5)²
  • x² - 2x + 1 + y² + 6y + 9 = 1/5
  • 5(x² + y² - 2x + 6y + 10) = 1

Final Answer: 5x² + 5y² - 10x + 30y + 49 = 0

UNIT-V: Conic Sections

Question 13 (c)

1 Mark

Question: Find the focus of the parabola y² = 16x.

Solution:

  • Comparing with y² = 4ax, we get 4a = 16, so a = 4.
  • Focus of y² = 4ax is (a, 0).

Final Answer: (4, 0)

Question 14 (a)

2 Marks

Question: Find the length of major axis and minor axis of the ellipse x²/25 + y²/16 = 1.

Solution:

  • Comparing with x²/a² + y²/b² = 1, we get a² = 25 (a = 5) and b² = 16 (b = 4).
  • Length of Major Axis = 2a = 2(5) = 10 units.
  • Length of Minor Axis = 2b = 2(4) = 8 units.

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