DISTANCE FORMULA - Questions & Answers

EXERCISE 25

1. Multiple Choice Type :
Choose the correct answer from the options given below.

(a) The distance between the points (7, -5) and (3, -1) is :
(i) 4 units     (ii) 3 units     (iii) 4√2 units     (iv) 5 units
Step 1: Use the distance formula: d = √[(x₂ - x₁)² + (y₂ - y₁)²]
Step 2: Substitute the given points: d = √[(3 - 7)² + (-1 - (-5))²]
Step 3: Simplify inside the brackets: d = √[(-4)² + (4)²]
Step 4: Square the numbers: d = √[16 + 16] = √32
Step 5: Simplify the square root: √32 = √(16 × 2) = 4√2 units.
Answer: (iii) 4√2 units

(b) The distance of point (-4, 3) from the origin is :
(i) 5 units     (ii) -5 units     (iii) 4 units     (iv) 3 units
Step 1: The origin is (0, 0). The distance formula from origin is d = √[x² + y²]
Step 2: Substitute the coordinates (-4, 3): d = √[(-4)² + (3)²]
Step 3: Square the numbers: d = √[16 + 9]
Step 4: Add the numbers: d = √25 = 5 units.
Answer: (i) 5 units

(c) The distance between the points (-3, 2) and (x, 10) is 10 units. The value of x is :
(i) 3     (ii) -9     (iii) 3 or -9     (iv) 3 and -9
Step 1: Use the distance formula and set it to 10: √[(x - (-3))² + (10 - 2)²] = 10
Step 2: Simplify inside: √[(x + 3)² + 8²] = 10
Step 3: Square both sides to remove the root: (x + 3)² + 64 = 100
Step 4: Subtract 64 from both sides: (x + 3)² = 36
Step 5: Take the square root of both sides: x + 3 = 6 or x + 3 = -6
Step 6: Solve for x: x = 3 or x = -9.
Answer: (iii) 3 or -9

(d) The point (x, y) is equidistant from the points (3, 6) and (-3, 4); the relation between x and y is :
(i) 3x - y = 5     (ii) 3x - y - 5 = 0     (iii) y - 3x = 0     (iv) 3x + y = 5
Step 1: Let P(x, y), A(3, 6), and B(-3, 4). Given PA = PB, so PA² = PB².
Step 2: Apply the distance formula squared: (x - 3)² + (y - 6)² = (x - (-3))² + (y - 4)²
Step 3: Expand both sides: x² - 6x + 9 + y² - 12y + 36 = x² + 6x + 9 + y² - 8y + 16
Step 4: Cancel x² and y² from both sides: -6x - 12y + 45 = 6x - 8y + 25
Step 5: Bring all terms to one side: -6x - 6x - 12y + 8y + 45 - 25 = 0
Step 6: Simplify: -12x - 4y + 20 = 0
Step 7: Divide the entire equation by -4: 3x + y - 5 = 0, which means 3x + y = 5.
Answer: (iv) 3x + y = 5

(e) The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is :
(i) 12 units     (ii) 6 units     (iii) 5 units     (iv) 10 units
Step 1: Let A(0, 4), B(0, 0), and C(3, 0). Perimeter = AB + BC + CA.
Step 2: Distance AB = √[(0 - 0)² + (0 - 4)²] = √16 = 4 units.
Step 3: Distance BC = √[(3 - 0)² + (0 - 0)²] = √9 = 3 units.
Step 4: Distance CA = √[(0 - 3)² + (4 - 0)²] = √[(-3)² + 4²] = √[9 + 16] = √25 = 5 units.
Step 5: Perimeter = 4 + 3 + 5 = 12 units.
Answer: (i) 12 units

2. Find the distance between the following pairs of points :
(i) (-3, 6) and (2, -6)
Step 1: Use distance formula: d = √[(x₂ - x₁)² + (y₂ - y₁)²]
Step 2: d = √[(2 - (-3))² + (-6 - 6)²]
Step 3: d = √[5² + (-12)²]
Step 4: d = √[25 + 144] = √169
Step 5: d = 13 units.
Answer: 13 units

(ii) (-a, -b) and (a, b)
Step 1: d = √[(a - (-a))² + (b - (-b))²]
Step 2: d = √[(2a)² + (2b)²]
Step 3: d = √[4a² + 4b²]
Step 4: Factor out 4: d = √[4(a² + b²)]
Step 5: d = 2√(a² + b²).
Answer: 2√(a² + b²)

(iii) (3/5, 2) and (-1/5, 1 2/5)
Step 1: Convert 1 2/5 to improper fraction: 7/5.
Step 2: d = √[(-1/5 - 3/5)² + (7/5 - 2)²]
Step 3: d = √[(-4/5)² + (7/5 - 10/5)²] = √[(-4/5)² + (-3/5)²]
Step 4: d = √[16/25 + 9/25] = √[25/25]
Step 5: d = √1 = 1 unit.
Answer: 1 unit

(iv) (√3 + 1, 1) and (0, √3)
Step 1: d = √[(0 - (√3 + 1))² + (√3 - 1)²]
Step 2: d = √[(-(√3 + 1))² + (√3 - 1)²]
Step 3: Expand: d = √[(3 + 2√3 + 1) + (3 - 2√3 + 1)]
Step 4: Combine like terms: d = √[4 + 2√3 + 4 - 2√3]
Step 5: d = √8 = 2√2 units.
Answer: 2√2 units

3. Find the distance between the origin and the point :
(i) (-8, 6)
Step 1: Distance from origin formula: d = √[x² + y²]
Step 2: d = √[(-8)² + 6²]
Step 3: d = √[64 + 36] = √100
Step 4: d = 10 units.
Answer: 10 units

(ii) (-5, -12)
Step 1: d = √[(-5)² + (-12)²]
Step 2: d = √[25 + 144] = √169
Step 3: d = 13 units.
Answer: 13 units

(iii) (8, -15)
Step 1: d = √[8² + (-15)²]
Step 2: d = √[64 + 225] = √289
Step 3: d = 17 units.
Answer: 17 units

4. The distance between the points (3, 1) and (0, x) is 5. Find x.
Step 1: Using distance formula: √[(0 - 3)² + (x - 1)²] = 5
Step 2: Square both sides: (-3)² + (x - 1)² = 25
Step 3: 9 + x² - 2x + 1 = 25
Step 4: x² - 2x + 10 = 25
Step 5: x² - 2x - 15 = 0
Step 6: Factor the quadratic equation: (x - 5)(x + 3) = 0
Step 7: x = 5 or x = -3.
Answer: x = 5 or x = -3

5. Find the co-ordinates of points on the x-axis which are at a distance of 17 units from the point (11, -8).
Step 1: Any point on the x-axis has y-coordinate 0. Let the point be (x, 0).
Step 2: Distance from (11, -8) is 17: √[(x - 11)² + (0 - (-8))²] = 17
Step 3: Square both sides: (x - 11)² + 8² = 289
Step 4: x² - 22x + 121 + 64 = 289
Step 5: x² - 22x + 185 - 289 = 0 => x² - 22x - 104 = 0
Step 6: Factorize: (x - 26)(x + 4) = 0
Step 7: x = 26 or x = -4.
Step 8: The coordinates are (26, 0) and (-4, 0).
Answer: (26, 0) and (-4, 0)

6. Find the co-ordinates of the points on the y-axis, which are at a distance of 10 units from the point (-8, 4).
Step 1: Any point on the y-axis has x-coordinate 0. Let the point be (0, y).
Step 2: Distance from (-8, 4) is 10: √[(0 - (-8))² + (y - 4)²] = 10
Step 3: Square both sides: 8² + (y - 4)² = 100
Step 4: 64 + y² - 8y + 16 = 100
Step 5: y² - 8y + 80 - 100 = 0 => y² - 8y - 20 = 0
Step 6: Factorize: (y - 10)(y + 2) = 0
Step 7: y = 10 or y = -2.
Step 8: The coordinates are (0, 10) and (0, -2).
Answer: (0, 10) and (0, -2)

7. A point A is at a distance of √10 units from the point (4, 3). Find the co-ordinates of point A, if its ordinate is twice its abscissa. (HOTS)
Step 1: Let the abscissa (x-coordinate) of A be x. The ordinate (y-coordinate) is twice the abscissa, so y = 2x. Let A = (x, 2x).
Step 2: Distance between (x, 2x) and (4, 3) is √10: √[(x - 4)² + (2x - 3)²] = √10
Step 3: Square both sides: (x - 4)² + (2x - 3)² = 10
Step 4: Expand: x² - 8x + 16 + 4x² - 12x + 9 = 10
Step 5: Combine terms: 5x² - 20x + 25 = 10
Step 6: Bring to one side: 5x² - 20x + 15 = 0
Step 7: Divide by 5: x² - 4x + 3 = 0
Step 8: Factorize: (x - 3)(x - 1) = 0 => x = 3 or x = 1.
Step 9: If x = 3, y = 6. If x = 1, y = 2.
Step 10: The coordinates are (3, 6) or (1, 2).
Answer: (3, 6) or (1, 2)

8. A point P (2, -1) is equidistant from the points (a, 7) and (-3, a). Find a.
Step 1: Distance from P(2, -1) to (a, 7) equals distance from P to (-3, a).
Step 2: √[(a - 2)² + (7 - (-1))²] = √[(-3 - 2)² + (a - (-1))²]
Step 3: Square both sides: (a - 2)² + 8² = (-5)² + (a + 1)²
Step 4: Expand: a² - 4a + 4 + 64 = 25 + a² + 2a + 1
Step 5: Cancel a² from both sides: -4a + 68 = 2a + 26
Step 6: Solve for a: 68 - 26 = 2a + 4a => 42 = 6a
Step 7: a = 7.
Answer: a = 7

9. What point on the x-axis is equidistant from the points (7, 6) and (-3, 4) ?
Step 1: Let the point on the x-axis be P(x, 0).
Step 2: Distance to (7, 6) equals distance to (-3, 4).
Step 3: (x - 7)² + (0 - 6)² = (x - (-3))² + (0 - 4)²
Step 4: x² - 14x + 49 + 36 = x² + 6x + 9 + 16
Step 5: Cancel x² and simplify: -14x + 85 = 6x + 25
Step 6: Solve for x: 85 - 25 = 6x + 14x => 60 = 20x => x = 3.
Step 7: The point is (3, 0).
Answer: (3, 0)

10. Find a point on the y-axis which is equidistant from the points (5, 2) and (-4, 3).
Step 1: Let the point on the y-axis be P(0, y).
Step 2: Distance to (5, 2) equals distance to (-4, 3).
Step 3: (0 - 5)² + (y - 2)² = (0 - (-4))² + (y - 3)²
Step 4: 25 + y² - 4y + 4 = 16 + y² - 6y + 9
Step 5: Cancel y² and simplify: -4y + 29 = -6y + 25
Step 6: Solve for y: 6y - 4y = 25 - 29 => 2y = -4 => y = -2.
Step 7: The point is (0, -2).
Answer: (0, -2)

11. A point P lies on the x-axis and another point Q lies on the y-axis.
(i) Write the ordinate of point P.
(ii) Write the abscissa of point Q.
(iii) If the abscissa of point P is -12 and the ordinate of point Q is -16; calculate the length of line segment PQ.
Step 1: (i) Any point on the x-axis has a y-coordinate (ordinate) of 0.
Step 2: (ii) Any point on the y-axis has an x-coordinate (abscissa) of 0.
Step 3: (iii) P has abscissa -12 and ordinate 0, so P = (-12, 0).
Step 4: Q has abscissa 0 and ordinate -16, so Q = (0, -16).
Step 5: Length PQ = √[(0 - (-12))² + (-16 - 0)²] = √[12² + (-16)²]
Step 6: PQ = √[144 + 256] = √400 = 20 units.
Answer: (i) 0    (ii) 0    (iii) 20 units

12. Show that the points P (0, 5), Q (5, 10) and R (6, 3) are the vertices of an isosceles triangle.
Step 1: Find distance PQ = √[(5 - 0)² + (10 - 5)²] = √[25 + 25] = √50 = 5√2 units.
Step 2: Find distance QR = √[(6 - 5)² + (3 - 10)²] = √[1² + (-7)²] = √[1 + 49] = √50 = 5√2 units.
Step 3: Find distance PR = √[(6 - 0)² + (3 - 5)²] = √[36 + (-2)²] = √[36 + 4] = √40 = 2√10 units.
Step 4: Since two sides are equal (PQ = QR), the points form an isosceles triangle.
Answer: PQ = QR = √50, hence proved it is an isosceles triangle.

13. Prove that the points P (0, -4), Q (6, 2), R (3, 5) and S (-3, -1) are the vertices of a rectangle PQRS.
Step 1: Find sides: PQ = √[(6-0)² + (2 - (-4))²] = √[36 + 36] = √72 units.
Step 2: QR = √[(3-6)² + (5-2)²] = √[(-3)² + 3²] = √[9 + 9] = √18 units.
Step 3: RS = √[(-3-3)² + (-1-5)²] = √[(-6)² + (-6)²] = √[36 + 36] = √72 units.
Step 4: SP = √[(0 - (-3))² + (-4 - (-1))²] = √[3² + (-3)²] = √[9 + 9] = √18 units.
Step 5: Opposite sides are equal (PQ = RS and QR = SP), so it's a parallelogram.
Step 6: Find diagonals: PR = √[(3-0)² + (5 - (-4))²] = √[9 + 81] = √90 units.
Step 7: QS = √[(-3-6)² + (-1-2)²] = √[(-9)² + (-3)²] = √[81 + 9] = √90 units.
Step 8: Since opposite sides are equal and diagonals are equal (PR = QS), PQRS is a rectangle.
Answer: Opposite sides are equal and diagonals are equal, hence proved it is a rectangle.

14. Prove that the points A (1, -3), B (-3, 0) and C (4, 1) are the vertices of an isosceles right-angled triangle. Find the area of the triangle.
Step 1: AB = √[(-3 - 1)² + (0 - (-3))²] = √[16 + 9] = √25 = 5 units.
Step 2: BC = √[(4 - (-3))² + (1 - 0)²] = √[49 + 1] = √50 units.
Step 3: CA = √[(1 - 4)² + (-3 - 1)²] = √[(-3)² + (-4)²] = √[9 + 16] = √25 = 5 units.
Step 4: Since AB = CA = 5, it is an isosceles triangle.
Step 5: Check Pythagoras theorem: AB² + CA² = 5² + 5² = 25 + 25 = 50. Also BC² = (√50)² = 50.
Step 6: Since AB² + CA² = BC², the triangle is right-angled at vertex A.
Step 7: Area = 1/2 × base × height = 1/2 × AB × CA = 1/2 × 5 × 5 = 25/2 = 12.5 square units.
Answer: Proved. Area = 12.5 sq units

15. Show that the points A (5, 6), B (1, 5), C (2, 1) and D (6, 2) are the vertices of a square ABCD.
Step 1: Find sides: AB = √[(1-5)² + (5-6)²] = √[16 + 1] = √17 units.
Step 2: BC = √[(2-1)² + (1-5)²] = √[1 + 16] = √17 units.
Step 3: CD = √[(6-2)² + (2-1)²] = √[16 + 1] = √17 units.
Step 4: DA = √[(5-6)² + (6-2)²] = √[1 + 16] = √17 units.
Step 5: All four sides are equal.
Step 6: Find diagonals: AC = √[(2-5)² + (1-6)²] = √[9 + 25] = √34 units.
Step 7: BD = √[(6-1)² + (2-5)²] = √[25 + 9] = √34 units.
Step 8: Since all sides are equal and diagonals are equal, ABCD is a square.
Answer: All sides equal (√17) and diagonals equal (√34), hence proved.

TEST YOURSELF

1. Multiple Choice Type :
Choose the correct answer from the options given below.

(a) The distance of point (-8, 6) from x-axis is :
(i) 8     (ii) 6     (iii) 10     (iv) none of these
Step 1: The distance of any point from the x-axis is the absolute value of its y-coordinate.
Step 2: The y-coordinate is 6, so distance is |6| = 6 units.
Answer: (ii) 6

(b) The distance of point (-4, -3) from the origin is :
(i) -10 unit     (ii) 10 unit     (iii) √[4² - 3²]     (iv) none of these
Step 1: Distance from origin = √[x² + y²]
Step 2: d = √[(-4)² + (-3)²] = √[16 + 9] = √25 = 5 units.
Step 3: Option (iii) evaluates to √ = √7, which is incorrect. None of the options matches exactly 5.
Answer: (iv) none of these

(c) The co-ordinates of point P are :
(i) (0, 5)     (ii) (5, 0)     (iii) (4, 3)     (iv) (3, 4)
Step 1: Observe the diagram. The right triangle has a base of 3 units on the x-axis, and hypotenuse of 5 units to point P.
Step 2: Use Pythagoras theorem to find the height (y-coordinate): Base² + Height² = Hypotenuse²
Step 3: 3² + y² = 5² => 9 + y² = 25 => y² = 16 => y = 4.
Step 4: The x-coordinate is 3. So the point P is (3, 4).
Answer: (iv) (3, 4)

(d) AB (= 10 unit) is diameter of a circle with centre at point P = (x, 0) and point B = (0, y). The relation between x and y is :
(i) x + y = 10     (ii) x + y = 25     (iii) x² + y² = 5     (iv) x² + y² = 25
Step 1: Radius of the circle is half the diameter, so radius = 10 / 2 = 5 units.
Step 2: The distance from centre P(x, 0) to point B(0, y) on the circle is the radius PB.
Step 3: PB = √[(0 - x)² + (y - 0)²] = 5
Step 4: Square both sides: x² + y² = 25.
Answer: (iv) x² + y² = 25

(e) Statement (1) : For the point P, x = -4 and y = 3, the distance of P from origin is 3 + 4 = 7.
Statement (2) : P = (-4, 3) and its distance from origin = √[(-4)² + 3²].
(i) Both the statements are true.
(ii) Both the statements are false.
(iii) Statement 1 is true, and statement 2 is false.
(iv) Statement 1 is false, and statement 2 is true.
Step 1: Check Statement 1: Distance = √[(-4)² + 3²] = √[16 + 9] = 5. Statement 1 claims distance is 7, so it is false.
Step 2: Check Statement 2: The formula shown √[(-4)² + 3²] is the correct distance formula from origin. So Statement 2 is true.
Answer: (iv) Statement 1 is false, and statement 2 is true.

(f) Statement (1) : The point P(x, y) is at a distance of 6 unit from origin, then P lies in the first quadrant.
Statement (2) : Point P can lie in any quadrant.
(i) Both the statements are true.
(ii) Both the statements are false.
(iii) Statement 1 is true, and statement 2 is false.
(iv) Statement 1 is false, and statement 2 is true.
Step 1: A point at a fixed distance from the origin forms a circle. The point can be in any of the four quadrants.
Step 2: Therefore, Statement 1 is false (it restricts P to the first quadrant only), and Statement 2 is true.
Answer: (iv) Statement 1 is false, and statement 2 is true.

(g) Assertion (A) : If A = (2x, y), B(x, 2y) and AB = 5 unit, then x + y = 5.
Reason (R) : √[(x - 2x)² + (2y - y)²] = 5 => x² + y² = 25
(i) A is true, R is false.
(ii) A is false, R is true.
(iii) Both A and R are true and R is the correct reason for A.
(iv) Both A and R are true and R is the incorrect reason for A.
Step 1: Evaluate Reason (R): Distance AB = √[(x - 2x)² + (2y - y)²] = √[(-x)² + y²] = √[x² + y²]. Given AB = 5, squaring gives x² + y² = 25. Reason (R) is true.
Step 2: Evaluate Assertion (A): x² + y² = 25 does NOT mean x + y = 5. (e.g., if x=3, y=4, x+y=7). Assertion (A) is false.
Answer: (ii) A is false, R is true.

(h) Assertion (A) : The distance between the points A(x, 2x) and B(x, 0) is 4 unit, the point B is (2, 0).
Reason (R) : √[(x - x)² + (2x - 0)²] = 4 => x² = 4 and x = ±2
(i) A is true, R is false.
(ii) A is false, R is true.
(iii) Both A and R are true and R is the correct reason for A.
(iv) Both A and R are true and R is the incorrect reason for A.
Step 1: Evaluate Reason (R): AB = √[(x - x)² + (0 - 2x)²] = √[(2x)²] = |2x|. Given AB = 4, so |2x| = 4 => 2x = ±4 => x = ±2. Squaring gives 4x²=16 => x²=4. Reason (R) is perfectly mathematically true.
Step 2: Evaluate Assertion (A): Since x can be 2 or -2, point B(x, 0) could be (2, 0) OR (-2, 0). The assertion states unequivocally "the point B is (2, 0)", which ignores the other valid point, making it technically false.
Answer: (ii) A is false, R is true.

2. Find the points on the y-axis which are at a distance of 2√5 units from the point (-4, 7).
Step 1: Let the point on the y-axis be P(0, y).
Step 2: Distance to (-4, 7) is 2√5: √[(0 - (-4))² + (y - 7)²] = 2√5
Step 3: Square both sides: 4² + (y - 7)² = (2√5)² = 4 × 5 = 20
Step 4: 16 + (y - 7)² = 20
Step 5: (y - 7)² = 4
Step 6: Take the square root: y - 7 = 2 or y - 7 = -2
Step 7: y = 9 or y = 5.
Step 8: The points are (0, 9) and (0, 5).
Answer: (0, 9) and (0, 5)

3. Find the value of k, if the points (5, k) and (k, 7) are equidistant from point (2, 4).
Step 1: Distance from (2, 4) to (5, k) = Distance from (2, 4) to (k, 7).
Step 2: √[(5 - 2)² + (k - 4)²] = √[(k - 2)² + (7 - 4)²]
Step 3: Square both sides: 3² + (k - 4)² = (k - 2)² + 3²
Step 4: Cancel 3² from both sides: (k - 4)² = (k - 2)²
Step 5: Expand: k² - 8k + 16 = k² - 4k + 4
Step 6: Cancel k² and bring k to one side: -8k + 4k = 4 - 16
Step 7: -4k = -12 => k = 3.
Answer: k = 3

4. The centre of a circle is (2a, a - 7). Find the value (values) of a, if the circle passes through the point (11, -9) and has diameter 10√2 units.
Step 1: Radius is half of the diameter, so radius = 10√2 / 2 = 5√2 units.
Step 2: The distance from centre (2a, a - 7) to point (11, -9) is the radius.
Step 3: √[(11 - 2a)² + (-9 - (a - 7))²] = 5√2
Step 4: Simplify inside the second bracket: (-9 - a + 7) = (-a - 2)
Step 5: Square both sides: (11 - 2a)² + (-a - 2)² = (5√2)² = 25 × 2 = 50
Step 6: Expand: 121 - 44a + 4a² + a² + 4a + 4 = 50
Step 7: Combine terms: 5a² - 40a + 125 = 50
Step 8: Bring to one side: 5a² - 40a + 75 = 0
Step 9: Divide by 5: a² - 8a + 15 = 0
Step 10: Factorize: (a - 5)(a - 3) = 0 => a = 5 or a = 3.
Answer: a = 5 or a = 3

5. Show that (-3, 2), (-5, -5), (2, -3) and (4, 4) are the vertices of a rhombus.
Step 1: Let A(-3, 2), B(-5, -5), C(2, -3), D(4, 4).
Step 2: Find AB = √[(-5 - (-3))² + (-5 - 2)²] = √[(-2)² + (-7)²] = √[4 + 49] = √53.
Step 3: Find BC = √[(2 - (-5))² + (-3 - (-5))²] = √[7² + 2²] = √[49 + 4] = √53.
Step 4: Find CD = √[(4 - 2)² + (4 - (-3))²] = √[2² + 7²] = √[4 + 49] = √53.
Step 5: Find DA = √[(-3 - 4)² + (2 - 4)²] = √[(-7)² + (-2)²] = √[49 + 4] = √53.
Step 6: Since all four sides are equal, it could be a rhombus or a square.
Step 7: Find diagonals to confirm. AC = √[(2 - (-3))² + (-3 - 2)²] = √[5² + (-5)²] = √50.
Step 8: BD = √[(4 - (-5))² + (4 - (-5))²] = √[9² + 9²] = √162.
Step 9: Since AC ≠ BD, the shape is a rhombus but not a square.
Answer: All sides equal, diagonals unequal, hence proved it is a rhombus.

6. Points A (-3, -2), B (-6, a), C (-3, -4) and D(0, -1) are the vertices of quadrilateral ABCD; find a if 'a' is negative and AB = CD.
Step 1: Find distance AB² = (-6 - (-3))² + (a - (-2))² = (-3)² + (a + 2)² = 9 + a² + 4a + 4 = a² + 4a + 13.
Step 2: Find distance CD² = (0 - (-3))² + (-1 - (-4))² = 3² + 3² = 9 + 9 = 18.
Step 3: Given AB = CD, so AB² = CD².
Step 4: a² + 4a + 13 = 18
Step 5: a² + 4a - 5 = 0
Step 6: Factorize: (a + 5)(a - 1) = 0 => a = -5 or a = 1.
Step 7: Since 'a' is negative, we discard 1.
Step 8: Therefore, a = -5.
Answer: a = -5

7. The vertices of a triangle are (5, 1), (11, 1) and (11, 9). Find the co-ordinates of the circumcentre of the triangle.
Step 1: Let the vertices be A(5, 1), B(11, 1) and C(11, 9). Let circumcentre be P(x, y).
Step 2: P is equidistant from all vertices: PA² = PB² = PC².
Step 3: PA² = PB² => (x - 5)² + (y - 1)² = (x - 11)² + (y - 1)²
Step 4: Cancel (y - 1)² from both sides: (x - 5)² = (x - 11)²
Step 5: x² - 10x + 25 = x² - 22x + 121 => 12x = 96 => x = 8.
Step 6: PB² = PC² => (x - 11)² + (y - 1)² = (x - 11)² + (y - 9)²
Step 7: Cancel (x - 11)² from both sides: (y - 1)² = (y - 9)²
Step 8: y² - 2y + 1 = y² - 18y + 81 => 16y = 80 => y = 5.
Step 9: The coordinates of the circumcentre are (8, 5).
Answer: (8, 5)

8. Given A = (3, 1) and B = (0, y - 1). Find y if AB = 5.
Step 1: Use distance formula squared: AB² = (0 - 3)² + ((y - 1) - 1)² = 5²
Step 2: (-3)² + (y - 2)² = 25
Step 3: 9 + (y - 2)² = 25
Step 4: (y - 2)² = 16
Step 5: Take square root: y - 2 = 4 or y - 2 = -4
Step 6: y = 6 or y = -2.
Answer: y = 6 or y = -2

9. Given A = (x + 2, -2) and B = (11, 6). Find x if AB = 17.
Step 1: Use distance formula squared: AB² = (11 - (x + 2))² + (6 - (-2))² = 17²
Step 2: (9 - x)² + 8² = 289
Step 3: (9 - x)² + 64 = 289
Step 4: (9 - x)² = 225
Step 5: Take square root: 9 - x = 15 or 9 - x = -15
Step 6: If 9 - x = 15 => x = -6. If 9 - x = -15 => x = 24.
Answer: x = -6 or x = 24

10. The centre of a circle is (2x - 1, 3x + 1). Find x if the circle passes through (-3, -1) and the length of its diameter is 20 units.
Step 1: Radius is diameter / 2 = 20 / 2 = 10 units.
Step 2: Distance from centre to point (-3, -1) is the radius, 10.
Step 3: Distance squared: (-3 - (2x - 1))² + (-1 - (3x + 1))² = 10²
Step 4: Simplify terms: (-2 - 2x)² + (-2 - 3x)² = 100
Step 5: Expand: (4 + 8x + 4x²) + (4 + 12x + 9x²) = 100
Step 6: Combine: 13x² + 20x + 8 = 100 => 13x² + 20x - 92 = 0
Step 7: Factorize: 13x² - 26x + 46x - 92 = 0 => 13x(x - 2) + 46(x - 2) = 0
Step 8: (13x + 46)(x - 2) = 0 => x = 2 or x = -46/13.
Answer: x = 2 or x = -46/13

11. The length of line PQ is 10 units and the co-ordinates of P are (2, -3); calculate the co-ordinates of point Q, if its abscissa is 10.
Step 1: Let point Q be (10, y) since its abscissa is 10.
Step 2: Distance PQ = 10. PQ² = (10 - 2)² + (y - (-3))² = 10²
Step 3: 8² + (y + 3)² = 100
Step 4: 64 + (y + 3)² = 100 => (y + 3)² = 36
Step 5: Take square root: y + 3 = 6 or y + 3 = -6
Step 6: y = 3 or y = -9.
Step 7: The coordinates of Q are (10, 3) or (10, -9).
Answer: (10, 3) or (10, -9)

12. Point P (2, -7) is the centre of a circle with radius 13 units, PT is perpendicular to chord AB and T = (-2, -4);
Calculate the length of : (i) AT   (ii) AB.
Step 1: First, find length of PT using distance formula between P(2, -7) and T(-2, -4).
Step 2: PT = √[(-2 - 2)² + (-4 - (-7))²] = √[(-4)² + 3²] = √25 = 5 units.
Step 3: In right-angled triangle PTA, hypotenuse PA is the radius (13 units).
Step 4: By Pythagoras theorem: PA² = PT² + AT²
Step 5: 13² = 5² + AT² => 169 = 25 + AT²
Step 6: AT² = 144 => AT = 12 units.
Step 7: A perpendicular from the centre bisects the chord, so AB = 2 × AT.
Step 8: AB = 2 × 12 = 24 units.
Answer: (i) 12 units   (ii) 24 units

13. Calculate the distance between the points P(2, 2) and Q(5, 4) correct to three significant figures.
Step 1: Distance PQ = √[(5 - 2)² + (4 - 2)²]
Step 2: PQ = √[3² + 2²]
Step 3: PQ = √[9 + 4] = √13.
Step 4: √13 ≈ 3.60555...
Step 5: Round correct to three significant figures: 3.61.
Answer: 3.61 units

14. Calculate the distance between A(7, 3) and B on the x-axis whose abscissa is 11.
Step 1: Point B is on the x-axis, so its y-coordinate is 0. B = (11, 0).
Step 2: Distance AB = √[(11 - 7)² + (0 - 3)²]
Step 3: AB = √[4² + (-3)²]
Step 4: AB = √[16 + 9] = √25.
Step 5: AB = 5 units.
Answer: 5 units

15. Calculate the distance between A(5, -3) and B on the y-axis whose ordinate is 9.
Step 1: Point B is on the y-axis, so its x-coordinate is 0. B = (0, 9).
Step 2: Distance AB = √[(0 - 5)² + (9 - (-3))²]
Step 3: AB = √[(-5)² + 12²]
Step 4: AB = √[25 + 144] = √169.
Step 5: AB = 13 units.
Answer: 13 units

16. Find the point on y-axis whose distances from the points A(6, 7) and B(4, -3) are in the ratio 1 : 2.
Step 1: Let point on y-axis be P(0, y). Given PA / PB = 1 / 2 => 2PA = PB => 4PA² = PB².
Step 2: Find PA² = (0 - 6)² + (y - 7)² = 36 + y² - 14y + 49 = y² - 14y + 85.
Step 3: Find PB² = (0 - 4)² + (y - (-3))² = 16 + y² + 6y + 9 = y² + 6y + 25.
Step 4: Substitute into 4PA² = PB²: 4(y² - 14y + 85) = y² + 6y + 25
Step 5: 4y² - 56y + 340 = y² + 6y + 25
Step 6: Bring to one side: 3y² - 62y + 315 = 0
Step 7: Factorize: 3y² - 27y - 35y + 315 = 0 => 3y(y - 9) - 35(y - 9) = 0
Step 8: (3y - 35)(y - 9) = 0 => y = 35/3 or y = 9.
Step 9: The points are (0, 35/3) and (0, 9).
Answer: (0, 35/3) and (0, 9)

17. The distances of point P(x, y) from the points A(1, -3) and B(-2, 2) are in the ratio 2 : 3.
Show that : 5x² + 5y² - 34x + 70y + 58 = 0. (HOTS)
Step 1: Given PA / PB = 2 / 3 => 3PA = 2PB => 9PA² = 4PB².
Step 2: PA² = (x - 1)² + (y - (-3))² = (x - 1)² + (y + 3)² = x² - 2x + 1 + y² + 6y + 9 = x² + y² - 2x + 6y + 10.
Step 3: PB² = (x - (-2))² + (y - 2)² = (x + 2)² + (y - 2)² = x² + 4x + 4 + y² - 4y + 4 = x² + y² + 4x - 4y + 8.
Step 4: Substitute into 9PA² = 4PB²: 9(x² + y² - 2x + 6y + 10) = 4(x² + y² + 4x - 4y + 8)
Step 5: Expand: 9x² + 9y² - 18x + 54y + 90 = 4x² + 4y² + 16x - 16y + 32
Step 6: Bring all to one side: (9x² - 4x²) + (9y² - 4y²) + (-18x - 16x) + (54y - (-16y)) + (90 - 32) = 0
Step 7: 5x² + 5y² - 34x + 70y + 58 = 0.
Answer: Hence Proved

18. The points A(3, 0), B(a, -2) and C(4, -1) are the vertices of triangle ABC right-angled at vertex A. Find the value of a.
Step 1: For right triangle at A, Pythagoras theorem applies: AB² + AC² = BC².
Step 2: AB² = (a - 3)² + (-2 - 0)² = a² - 6a + 9 + 4 = a² - 6a + 13.
Step 3: AC² = (4 - 3)² + (-1 - 0)² = 1² + (-1)² = 1 + 1 = 2.
Step 4: BC² = (4 - a)² + (-1 - (-2))² = 16 - 8a + a² + 1² = a² - 8a + 17.
Step 5: Substitute into equation: (a² - 6a + 13) + 2 = a² - 8a + 17
Step 6: a² - 6a + 15 = a² - 8a + 17
Step 7: Cancel a² and solve for a: 8a - 6a = 17 - 15 => 2a = 2 => a = 1.
Answer: a = 1

19. If two vertices of an equilateral triangle be (0, 0) and (3, √3), find the third vertex using distance formula.
Step 1: Let the third vertex be P(x, y). Let A(0, 0) and B(3, √3).
Step 2: In equilateral triangle, AB = PA = PB. AB² = (3 - 0)² + (√3 - 0)² = 9 + 3 = 12.
Step 3: PA² = x² + y² = 12. (Equation 1)
Step 4: PB² = (x - 3)² + (y - √3)² = 12.
Step 5: Expand PB²: x² - 6x + 9 + y² - 2√3y + 3 = 12. Grouping gives (x² + y²) - 6x - 2√3y + 12 = 12.
Step 6: Substitute x² + y² = 12 into this: 12 - 6x - 2√3y + 12 = 12 => 12 - 6x - 2√3y = 0.
Step 7: Simplify: 6x + 2√3y = 12 => 3x + √3y = 6 => √3y = 6 - 3x => y = √3(2 - x).
Step 8: Substitute y in Equation 1: x² + (√3(2 - x))² = 12 => x² + 3(4 - 4x + x²) = 12.
Step 9: x² + 12 - 12x + 3x² = 12 => 4x² - 12x = 0 => 4x(x - 3) = 0.
Step 10: x = 0 or x = 3.
Step 11: If x = 0, y = √3(2 - 0) = 2√3. If x = 3, y = √3(2 - 3) = -√3.
Step 12: The third vertex can be (0, 2√3) or (3, -√3).
Answer: (0, 2√3) or (3, -√3)

Case-Study Based Question

1. A group of class IX students goes to a picnic during winter holidays. The positions of three friends Nitin, Rajesh and Kareem are shown by the points P, Q and R respectively.
(i) Find the distance between :
(a) Nitin and Rajesh
(b) Rajesh and Kareem
(c) Nitin and Kareem
Step 1: From the grid graph, identify coordinates. Nitin (P) is at (6, 4). Kareem (R) is at (9, 7). Rajesh (Q) is at (11, 9).
Step 2: (a) Distance P (Nitin) to Q (Rajesh) = √[(11 - 6)² + (9 - 4)²] = √[5² + 5²] = √50 = 5√2 units.
Step 3: (b) Distance Q (Rajesh) to R (Kareem) = √[(11 - 9)² + (9 - 7)²] = √[2² + 2²] = √8 = 2√2 units.
Step 4: (c) Distance P (Nitin) to R (Kareem) = √[(9 - 6)² + (7 - 4)²] = √[3² + 3²] = √18 = 3√2 units.
Answer: (a) 5√2 units (b) 2√2 units (c) 3√2 units

(ii) Show that P, Q and R are collinear.
Step 1: For points to be collinear, the sum of distances of two smaller segments must equal the largest segment.
Step 2: Here, PR = 3√2, RQ = 2√2, and PQ = 5√2.
Step 3: Check: PR + RQ = 3√2 + 2√2 = 5√2.
Step 4: Since PR + RQ = PQ, the points lie on the same straight line.
Answer: PR + RQ = PQ, hence proved they are collinear.

(iii) Find the point on X-axis which is equidistant from points Q and R.
Step 1: Let point on x-axis be S(x, 0). It is equidistant from Q(11, 9) and R(9, 7). SQ² = SR².
Step 2: (x - 11)² + (0 - 9)² = (x - 9)² + (0 - 7)²
Step 3: x² - 22x + 121 + 81 = x² - 18x + 81 + 49
Step 4: Cancel x² and simplify: -22x + 202 = -18x + 130
Step 5: 202 - 130 = 22x - 18x => 72 = 4x => x = 18.
Step 6: The point is (18, 0).
Answer: (18, 0)

(iv) If R is taken as origin; what will be the co-ordinates of P and Q ?
Step 1: The new origin is R(9, 7). To find new coordinates, subtract origin coordinates from old coordinates.
Step 2: New P = (6 - 9, 4 - 7) = (-3, -3).
Step 3: New Q = (11 - 9, 9 - 7) = (2, 2).
Answer: P(-3, -3) and Q(2, 2)

2. Rohan and Sohan stay in a village which are 50 km apart and are situated on Delhi Agra Highway as shown in the figure. Another highway YY' crosses Agra Delhi highway at O (0, 0). A small local road PQ crosses both the highways at points A and B such that OA = 10 km and OB = 12 km. Also the villages of Sita and Gita are on the smaller highway YY'. Sita's village is 12 km from O and Gita's village is 15 km from O.
Now answer the following questions:
(i) Find the Coordinates of A and B.
Step 1: Looking at the diagram, A is on the positive x-axis (Agra Delhi highway), 10 km from O. So A = (10, 0).
Step 2: B is on the positive y-axis (smaller highway), 12 km from O. So B = (0, 12).
Answer: A(10, 0) and B(0, 12)

(ii) Find the Length of local road AB.
Step 1: Distance AB = √[(10 - 0)² + (0 - 12)²]
Step 2: AB = √[10² + (-12)²] = √[100 + 144]
Step 3: AB = √244 = √(4 × 61) = 2√61 km.
Answer: 2√61 km

(iii) What is the shortest distance Sohan will cover to reach Sita's village.
Step 1: Form map: Sohan is 20 km right of A(10, 0), so Sohan = (30, 0). Sita is at B(0, 12) because her village is 12 km from O on +ve Y axis.
Step 2: Shortest distance is the straight line from (30, 0) to (0, 12).
Step 3: d = √[(30 - 0)² + (0 - 12)²]
Step 4: d = √[900 + 144] = √1044 = √(36 × 29) = 6√29 km.
Answer: 6√29 km

(iv) How far is Rohan's village from Gita's village?
Step 1: Rohan is 20 km to the left of O, so Rohan = (-20, 0). Gita is 15 km below O, so Gita = (0, -15).
Step 2: Distance = √[(-20 - 0)² + (0 - (-15))²]
Step 3: d = √[(-20)² + 15²]
Step 4: d = √[400 + 225] = √625 = 25 km.
Answer: 25 km