CBSE Class 10 Mathematics Coordinate Geometry Assignment Set 22

Exercise I: Distance Formula

Question 1. Find the distance between the following pairs of points: (i) \( (2, 3),\ (4, 1) \); (ii) \( (-5, 7),\ (-1, 3) \); (iii) \( (a, b),\ (-a, -b) \).
Answer:
(i) \( PQ = \sqrt{(4 - 2)^2 + (1 - 3)^2} = \sqrt{4 + 4} = \) \( 2\sqrt{2} \) units.
(ii) \( AB = \sqrt{(-1 + 5)^2 + (3 - 7)^2} = \sqrt{16 + 16} = \) \( 4\sqrt{2} \) units.
(iii) \( CK = \sqrt{(-a - a)^2 + (-b - b)^2} = \sqrt{4a^2 + 4b^2} = \) \( 2\sqrt{a^2 + b^2} \) units.
In simple words: The distance is the square root of the sum of the squares of the differences of the x-coordinates and of the y-coordinates.

Teacher's Note:
1. Write the two points as (x₁, y₁) and (x₂, y₂), substitute in the distance formula and simplify the root.
2. Leave the answer in the simplest surd form.

Question 2. Find the distance between the points \( (0, 0) \) and \( (36, 15) \). Can you now find the distance between the two towns A and B?
The map shows town A at the origin and town B 36 km east and 15 km north of A.
Answer:
\( AB = \sqrt{(36 - 0)^2 + (15 - 0)^2} = \sqrt{1296 + 225} = \sqrt{1521} = 39 \).
Hence the distance between the two towns is 39 km.
In simple words: Town A is at the origin and town B is 36 km east and 15 km north of it, so the straight-line distance follows from the distance formula.

Teacher's Note:
1. Write the two points as (x₁, y₁) and (x₂, y₂), substitute in the distance formula and simplify the root.
2. Leave the answer in the simplest surd form.

Question 3. Determine if the points \( (1, 5),\ (2, 3) \) and \( (-2, -11) \) are collinear.
Answer:
Let \( A(1, 5),\ B(2, 3),\ C(-2, -11) \).
\( AB = \sqrt{(2 - 1)^2 + (3 - 5)^2} = \sqrt{5} \), \( BC = \sqrt{(-2 - 2)^2 + (-11 - 3)^2} = \sqrt{212} = 2\sqrt{53} \), \( CA = \sqrt{(-2 - 1)^2 + (-11 - 5)^2} = \sqrt{265} = \sqrt{5}\sqrt{53} \).
Here \( AB + BC = \sqrt{5} + 2\sqrt{53} \approx 16.8 \neq CA \approx 16.3 \), and no other sum of two sides equals the third.
Hence A, B and C are not collinear.
In simple words: Three points are collinear only if the two shorter distances add up to the longest one.

Teacher's Note:
1. Compute all three distances before comparing.
2. Compare the sum of the two smaller distances with the largest one.

Question 4. Check whether \( (5, -2),\ (6, 4) \) and \( (7, -2) \) are the vertices of an isosceles triangle.

Answer:
Let \( A(5, -2),\ B(6, 4),\ C(7, -2) \).
\( AB = \sqrt{(6 - 5)^2 + (4 + 2)^2} = \sqrt{37} \), \( BC = \sqrt{(7 - 6)^2 + (-2 - 4)^2} = \sqrt{37} \), \( AC = \sqrt{(7 - 5)^2 + 0} = 2 \).
Since \( AB = BC \), \( \triangle ABC \) is an isosceles triangle.
In simple words: Two equal sides make the triangle isosceles.

Teacher's Note:
1. Find all three side lengths.
2. An isosceles triangle has at least two equal sides.

Question 5. In a classroom, 4 friends are seated at the points A, B, C and D as shown in the figure. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, \"Don't you think ABCD is a square?\" Chameli disagrees. Using distance formula, find which of them is correct.

Answer:
From the figure, \( A(3, 4),\ B(6, 7),\ C(9, 4),\ D(6, 1) \).
\( AB = \sqrt{(6 - 3)^2 + (7 - 4)^2} = 3\sqrt{2} \), \( BC = \sqrt{(9 - 6)^2 + (4 - 7)^2} = 3\sqrt{2} \), \( CD = \sqrt{(6 - 9)^2 + (1 - 4)^2} = 3\sqrt{2} \), \( DA = \sqrt{(6 - 3)^2 + (1 - 4)^2} = 3\sqrt{2} \).
The diagonals are \( AC = \sqrt{(9 - 3)^2 + 0} = 6 \) and \( BD = \sqrt{0 + (1 - 7)^2} = 6 \).
All four sides are equal and the diagonals are equal, so ABCD is a square and Champa is correct.
In simple words: A quadrilateral with four equal sides and two equal diagonals is a square.

Teacher's Note:
1. Equal sides alone would also fit a rhombus, so the diagonals must be checked.
2. Check both diagonals to separate a square from a rhombus.

Question 6. Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (i) \( (-1, -2),\ (1, 0),\ (-1, 2),\ (-3, 0) \); (ii) \( (-3, 5),\ (3, 1),\ (0, 3),\ (-1, -4) \); (iii) \( (4, 5),\ (7, 6),\ (4, 3),\ (1, 2) \).
Answer:
(i) Let \( A(-1, -2),\ B(1, 0),\ C(-1, 2),\ D(-3, 0) \). \( AB = BC = CD = DA = 2\sqrt{2} \) and the diagonals \( AC = BD = 4 \). All sides and both diagonals are equal, so it is a square.
(ii) Let \( A(-3, 5),\ B(3, 1),\ C(0, 3),\ D(-1, -4) \). \( AB = \sqrt{36 + 16} = 2\sqrt{13} \), \( AC = \sqrt{9 + 4} = \sqrt{13} \) and \( CB = \sqrt{9 + 4} = \sqrt{13} \). Here \( AC + CB = AB \), so A, C and B are collinear (C is the mid-point of AB). Three of the four points lie on one line, so no quadrilateral is formed.
(iii) Let \( A(4, 5),\ B(7, 6),\ C(4, 3),\ D(1, 2) \). \( AB = CD = \sqrt{10} \) and \( BC = DA = 3\sqrt{2} \), while the diagonals \( AC = 2 \) and \( BD = 2\sqrt{13} \) are not equal. Opposite sides are equal and the diagonals are unequal, so it is a parallelogram.
In simple words: Compare the four sides and the two diagonals: equal sides with equal diagonals give a square, equal opposite sides with unequal diagonals give a parallelogram.

Teacher's Note:
1. Before naming the quadrilateral, check that no three of the points are collinear.
2. Square: four equal sides and equal diagonals; parallelogram: opposite sides equal and diagonals unequal.

Question 7. Find the point on the x-axis which is equidistant from \( (2, -5) \) and \( (-2, 9) \).
Answer:
Let the point on the x-axis be \( R(x, 0) \), \( P(2, -5) \) and \( Q(-2, 9) \). Then \( RP = RQ \Rightarrow RP^2 = RQ^2 \).
\( (2 - x)^2 + 25 = (-2 - x)^2 + 81 \Rightarrow 4 - 4x + x^2 + 25 = 4 + 4x + x^2 + 81 \Rightarrow -8x = 56 \Rightarrow x = -7 \).
Hence the required point is \( (-7, 0) \).
In simple words: A point on the x-axis has the form (x, 0); equate the squared distances to the two points.

Teacher's Note:
1. Squaring both sides removes the square roots.
2. The x² terms cancel, leaving a linear equation.

Question 8. Find the values of y for which the distance between the points \( P(2, -3) \) and \( Q(10, y) \) is 10 units.
Answer:
\( PQ = \sqrt{(10 - 2)^2 + (y + 3)^2} = 10 \Rightarrow 64 + (y + 3)^2 = 100 \Rightarrow (y + 3)^2 = 36 \Rightarrow y + 3 = \pm 6 \).
Hence \( y = 3 \) or \( y = -9 \).
In simple words: Square both sides of the distance equation and solve the resulting quadratic.

Teacher's Note:
1. Do not forget the negative square root.
2. Both values of y satisfy the condition.

Question 9. If \( Q(0, 1) \) is equidistant from \( P(5, -3) \) and \( R(x, 6) \), find the values of x. Also find the distances QR and PR.
Answer:
\( QP = \sqrt{25 + 16} = \sqrt{41} \) and \( QR = \sqrt{x^2 + 25} \). Since \( QP = QR \), \( 41 = x^2 + 25 \Rightarrow x^2 = 16 \Rightarrow x = \pm 4 \).
So R is \( (4, 6) \) or \( (-4, 6) \).
If \( R(4, 6) \): \( QR = \sqrt{16 + 25} = \sqrt{41} \) and \( PR = \sqrt{(5 - 4)^2 + (-3 - 6)^2} = \sqrt{82} \).
If \( R(-4, 6) \): \( QR = \sqrt{41} \) and \( PR = \sqrt{(5 + 4)^2 + (-3 - 6)^2} = \sqrt{162} = 9\sqrt{2} \).
Hence \( x = \pm 4 \); \( QR = \sqrt{41} \) units; \( PR = \sqrt{82} \) units (for \( x = 4 \)) or \( 9\sqrt{2} \) units (for \( x = -4 \)).
In simple words: Equal distances give x² = 16, so there are two possible positions for R.

Teacher's Note:
1. Treat the two positions of R separately when finding PR.
2. The value of QR is the same for both positions.

Question 10. Find a relation between x and y such that the point \( (x, y) \) is equidistant from the points \( (3, 6) \) and \( (-3, 4) \).
Answer:
Let \( P(x, y) \) be equidistant from \( A(3, 6) \) and \( B(-3, 4) \). Then \( PA^2 = PB^2 \).
\( (x - 3)^2 + (y - 6)^2 = (x + 3)^2 + (y - 4)^2 \Rightarrow -6x - 12y + 45 = 6x - 8y + 25 \Rightarrow 12x + 4y = 20 \).
Hence the required relation is \( 3x + y - 5 = 0 \).
In simple words: The points equidistant from two given points lie on the perpendicular bisector of the segment joining them.

Teacher's Note:
1. Expand both squared distances carefully.
2. Cancel the x² and y² terms from both sides.