Coordinate Geometry - Distance & Section Formula for Exams

Coordinate Geometry - Distance and Section Formula - Problems on Circles - Applications

Q. A point Q divides the segment joining P(1, 1) and R(7, 5) in the ratio 2:1. What are the coordinates of Q?

A. (5, 3)
B. (4, 2)
C. (6, 4)
D. (3, 2)

Solution

Using the section formula: Q = ((mx2 + nx1)/(m+n), (my2 + ny1)/(m+n)) where m=2, n=1. Q = ((2*7 + 1*1)/(2+1), (2*5 + 1*1)/(2+1)) = (5, 3).

Correct Answer: A — (5, 3)

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Q. Find the coordinates of the point that divides the segment joining (1, 2) and (3, 4) in the ratio 1:3.

A. (2, 3)
B. (1.5, 2.5)
C. (2.5, 3.5)
D. (3, 4)

Solution

Using the section formula: (mx2 + nx1)/(m+n), (my2 + ny1)/(m+n) = (1*3 + 3*1)/(1+3), (1*4 + 3*2)/(1+3) = (3 + 3)/4, (4 + 6)/4 = (6/4, 10/4) = (1.5, 2.5).

Correct Answer: C — (2.5, 3.5)

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Q. Find the midpoint of the line segment joining the points (1, 2) and (5, 6).

A. (3, 4)
B. (4, 3)
C. (2, 5)
D. (5, 2)

Solution

Midpoint M = ((x1 + x2)/2, (y1 + y2)/2) = ((1 + 5)/2, (2 + 6)/2) = (3, 4).

Correct Answer: A — (3, 4)

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Q. If a circle has a center at (0, 0) and passes through the point (3, 4), what is its radius?

A. 5
B. 7
C. 4
D. 3

Solution

Radius r = distance from center to point = √((3 - 0)² + (4 - 0)²) = √(9 + 16) = √25 = 5.

Correct Answer: A — 5

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Q. If a circle has a diameter of 10 units, what is its circumference?

A. 10π
B. 20π
C. 5π
D. 15π

Solution

Circumference = πd = π(10) = 10π units.

Correct Answer: B — 20π

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Q. If a point P divides the line segment joining A(2, 3) and B(8, 7) in the ratio 1:3, what are the coordinates of point P?

A. (5, 5)
B. (4, 4)
C. (6, 6)
D. (3, 3)

Solution

Using the section formula: P = ((mx2 + nx1)/(m+n), (my2 + ny1)/(m+n)) where m=1, n=3, A(2,3), B(8,7). P = ((1*8 + 3*2)/(1+3), (1*7 + 3*3)/(1+3)) = (5, 5).

Correct Answer: A — (5, 5)

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Q. If the coordinates of the center of a circle are (4, -2) and it passes through the point (4, 2), what is the radius of the circle?

A. 4
B. 2
C. 6
D. 8

Solution

Using the distance formula: radius = √((4 - 4)² + (2 - (-2))²) = √(0 + 16) = √16 = 4.

Correct Answer: B — 2

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Q. If the coordinates of the vertices of a triangle are A(1, 1), B(4, 5), and C(7, 2), what is the length of side AB?

A. 5
B. 4
C. 3
D. 6

Solution

Length AB = √((4 - 1)² + (5 - 1)²) = √(9 + 16) = √25 = 5.

Correct Answer: A — 5

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Q. What is the distance from the point (2, 3) to the line 2x + 3y - 6 = 0?

A. 1
B. 2
C. 3
D. 4

Solution

Distance = |Ax1 + By1 + C| / √(A² + B²) = |2(2) + 3(3) - 6| / √(2² + 3²) = |4 + 9 - 6| / √13 = 7 / √13 ≈ 1.94, which rounds to 2.

Correct Answer: B — 2

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Q. What is the equation of a circle with center at (0, 0) and radius 3?

A. x² + y² = 9
B. x² + y² = 3
C. x² + y² = 6
D. x² + y² = 12

Solution

The standard form is (x - h)² + (y - k)² = r². Here, h=0, k=0, r=3, so the equation is x² + y² = 3² = 9.

Correct Answer: A — x² + y² = 9

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Q. What is the equation of a circle with center at (0, 0) and radius 4?

A. x² + y² = 16
B. x² + y² = 8
C. x² + y² = 4
D. x² + y² = 20

Solution

The equation of a circle is (x - h)² + (y - k)² = r². Here, h = 0, k = 0, r = 4, so x² + y² = 4² = 16.

Correct Answer: A — x² + y² = 16

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Q. What is the length of the diameter of a circle with a radius of 7 units?

A. 14
B. 7
C. 21
D. 28

Solution

Diameter d = 2 * radius = 2 * 7 = 14 units.

Correct Answer: A — 14

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Q. What is the length of the line segment joining the points (1, 1) and (4, 5)?

A. 4
B. 5
C. 3
D. 6

Solution

Using the distance formula: d = √((4 - 1)² + (5 - 1)²) = √(9 + 16) = √25 = 5.

Correct Answer: B — 5

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Q. What is the radius of a circle with the equation (x - 2)² + (y + 3)² = 16?

A. 4
B. 8
C. 2
D. 3

Solution

The standard form of a circle is (x - h)² + (y - k)² = r². Here, r² = 16, so r = √16 = 4.

Correct Answer: A — 4

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Showing 1 to 14 of 14 (1 Pages)

Coordinate Geometry - Distance and Section Formula - Problems on Circles - Applications

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