Selina Concise Solutions for ICSE Class 9 Mathematics Chapter 28 Distance Formula

Exercise 1(A)

Question 1. Find the distance between the following pairs of points:
(i) \( (-3, 6) \) and \( (2, -6) \)
(ii) \( (-a, -b) \) and \( (a, b) \)
(iii) \( \left(\frac{3}{5}, 2\right) \) and \( \left(-\frac{1}{5}, 1\frac{2}{5}\right) \)
(iv) \( (\sqrt{3} + 1, 1) \) and \( (0, \sqrt{3}) \)
Answer:
(i) Using the distance formula, we calculate the distance between \( (-3, 6) \) and \( (2, -6) \):
Distance \( = \sqrt{(2 - (-3))^2 + (-6 - 6)^2} \)
\( = \sqrt{(5)^2 + (-12)^2} \)
\( = \sqrt{25 + 144} \)
\( = \sqrt{169} = 13 \)

(ii) Calculating the distance between \( (-a, -b) \) and \( (a, b) \):
Distance \( = \sqrt{(a - (-a))^2 + (b - (-b))^2} \)
\( = \sqrt{(2a)^2 + (2b)^2} \)
\( = \sqrt{4a^2 + 4b^2} \)
\( = 2\sqrt{a^2 + b^2} \)

(iii) Determining the distance between \( \left(\frac{3}{5}, 2\right) \) and \( \left(-\frac{1}{5}, 1\frac{2}{5}\right) \):
Distance \( = \sqrt{\left(-\frac{1}{5} - \frac{3}{5}\right)^2 + \left(1\frac{2}{5} - 2\right)^2} \)
\( = \sqrt{\left(-\frac{4}{5}\right)^2 + \left(\frac{7}{5} - \frac{10}{5}\right)^2} \)
\( = \sqrt{\left(-\frac{4}{5}\right)^2 + \left(-\frac{3}{5}\right)^2} \)
\( = \sqrt{\frac{16}{25} + \frac{9}{25}} \)
\( = \sqrt{\frac{25}{25}} = 1 \)

(iv) Finding the distance between \( (\sqrt{3} + 1, 1) \) and \( (0, \sqrt{3}) \):
Distance \( = \sqrt{(0 - (\sqrt{3} + 1))^2 + (\sqrt{3} - 1)^2} \)
\( = \sqrt{(-\sqrt{3} - 1)^2 + (\sqrt{3} - 1)^2} \)
\( = \sqrt{(\sqrt{3} + 1)^2 + (\sqrt{3} - 1)^2} \)
\( = \sqrt{3 + 1 + 2\sqrt{3} + 3 + 1 - 2\sqrt{3}} \)
\( = \sqrt{8} = 2\sqrt{2} \)
In simple words: To find the distance between any two coordinate points, we subtract the x-values and square the result, do the same for the y-values, add them together, and then find the square root of that sum.

Exam Tip: Be very careful with signs when substituting coordinates into the distance formula. For instance, subtracting a negative coordinate like -3 becomes adding 3.

Question 2. Find the distance between the origin and the following points:
(i) \( A(-8, 6) \)
(ii) \( B(-5, -12) \)
(iii) \( C(8, -15) \)
Answer:
Let the origin be represented by the coordinates \( O(0, 0) \).
(i) For point \( A(-8, 6) \), the distance is:
\( AO = \sqrt{(0 - (-8))^2 + (0 - 6)^2} \)
\( = \sqrt{8^2 + (-6)^2} \)
\( = \sqrt{64 + 36} \)
\( = \sqrt{100} = 10 \) units

(ii) For point \( B(-5, -12) \), we have:
\( BO = \sqrt{(0 - (-5))^2 + (0 - (-12))^2} \)
\( = \sqrt{5^2 + 12^2} \)
\( = \sqrt{25 + 144} \)
\( = \sqrt{169} = 13 \) units

(iii) For point \( C(8, -15) \), we get:
\( CO = \sqrt{(0 - 8)^2 + (0 - (-15))^2} \)
\( = \sqrt{(-8)^2 + 15^2} \)
\( = \sqrt{64 + 225} \)
\( = \sqrt{289} = 17 \) units
In simple words: The distance of any coordinate point from the origin is calculated by squaring both coordinates, adding them, and taking the square root.

Exam Tip: The formula for distance from the origin simplifies to \( d = \sqrt{x^2 + y^2} \). Remember that squaring a negative number always yields a positive result.

Question 3. Find the value of \( x \) if the distance between the points \( A(3, 1) \) and \( B(0, x) \) is 5.
Answer:
We are given that the distance between \( A(3, 1) \) and \( B(0, x) \) is equal to 5.
This means:
\( AB = 5 \)

\( \implies AB^2 = 25 \)
Using the distance formula:
\( (0 - 3)^2 + (x - 1)^2 = 25 \)
\( 9 + x^2 - 2x + 1 = 25 \)
\( x^2 - 2x + 10 - 25 = 0 \)
\( x^2 - 2x - 15 = 0 \)
Splitting the middle term:
\( x^2 - 5x + 3x - 15 = 0 \)
\( x(x - 5) + 3(x - 5) = 0 \)
\( (x - 5)(x + 3) = 0 \)
Thus, we find:
\( x = 5 \) or \( x = -3 \)
In simple words: We use the distance formula to set up a quadratic equation. Solving this equation gives us the possible values for the unknown coordinate.

Exam Tip: When solving quadratic equations in coordinate geometry, do not discard negative values unless the question explicitly restricts the domain (e.g., stating \( x \) must be positive).

Question 4. Find the coordinates of the points on the x-axis which are at a distance of 17 units from the point \((11, -8)\).
Answer:
Since the required point lies on the x-axis, its coordinates can be represented as \( (x, 0) \).
According to the problem, the distance from this point to \((11, -8)\) is 17 units.
Applying the distance formula:
\( \sqrt{(x - 11)^2 + (0 - (-8))^2} = 17 \)

\( \implies (x - 11)^2 + 8^2 = 289 \)
\( x^2 - 22x + 121 + 64 = 289 \)
\( x^2 - 22x + 185 - 289 = 0 \)
\( x^2 - 22x - 104 = 0 \)
Let us factor this quadratic equation:
\( x^2 - 26x + 4x - 104 = 0 \)
\( x(x - 26) + 4(x - 26) = 0 \)
\( (x - 26)(x + 4) = 0 \)
This gives:
\( x = 26 \) or \( x = -4 \)
Therefore, the coordinates of the points on the x-axis are \((26, 0)\) and \((-4, 0)\).
In simple words: Any point on the x-axis has a y-coordinate of 0. We substitute this into the distance formula with the given point and solve for the unknown x-coordinate.

Exam Tip: Always write your final answer as coordinate pairs, such as \((26, 0)\) and \((-4, 0)\), rather than just leaving the values as \( x = 26, -4 \).

Question 5. Find the coordinates of the points on the y-axis which are at a distance of 10 units from the point \((-8, 4)\).
Answer:
Any point on the y-axis has coordinates in the form of \( (0, y) \).
The distance between \( (0, y) \) and the given point \((-8, 4)\) is specified as 10 units.
Using the distance formula:
\( \sqrt{(0 - (-8))^2 + (y - 4)^2} = 10 \)
Squaring both sides of the equation:
\( (8)^2 + (y - 4)^2 = 100 \)
\( 64 + y^2 - 8y + 16 = 100 \)
\( y^2 - 8y + 80 - 100 = 0 \)
\( y^2 - 8y - 20 = 0 \)
Factoring the expression:
\( y^2 - 10y + 2y - 20 = 0 \)
\( y(y - 10) + 2(y - 10) = 0 \)
\( (y - 10)(y + 2) = 0 \)
This yields:
\( y = 10 \) or \( y = -2 \)
Hence, the coordinates of the points on the y-axis are \((0, 10)\) and \((0, -2)\).
In simple words: A point on the y-axis has an x-coordinate of 0. We set the distance equation to 10 and solve the resulting quadratic equation to find the two possible coordinates.

Exam Tip: Remember to write the coordinates with 0 as the x-component, as the question specifies that the points are on the y-axis.

Question 6. A point A is at a distance of \(\sqrt{10}\) units from the point \((4, 3)\). Find the coordinates of point A, if its ordinate is twice its abscissa.
Answer:
We are given that the ordinate (y-coordinate) of point A is twice its abscissa (x-coordinate).
Let the coordinates of point A be \( (x, 2x) \).
The distance of this point A from \((4, 3)\) is given as \(\sqrt{10}\) units.
Applying the distance formula:
\( \sqrt{(x - 4)^2 + (2x - 3)^2} = \sqrt{10} \)
Squaring both sides:
\( (x - 4)^2 + (2x - 3)^2 = 10 \)
\( x^2 - 8x + 16 + 4x^2 - 12x + 9 = 10 \)
\( 5x^2 - 20x + 25 - 10 = 0 \)
\( 5x^2 - 20x + 15 = 0 \)
Dividing the entire equation by 5:
\( x^2 - 4x + 3 = 0 \)
Factorizing:
\( x^2 - x - 3x + 3 = 0 \)
\( x(x - 1) - 3(x - 1) = 0 \)
\( (x - 1)(x - 3) = 0 \)
Thus, the values are:
\( x = 1 \) or \( x = 3 \)
- If \( x = 1 \), the coordinates are \((1, 2(1)) = (1, 2)\).
- If \( x = 3 \), the coordinates are \((3, 2(3)) = (3, 6)\).
Hence, the coordinates of point A can be \((1, 2)\) or \((3, 6)\).
In simple words: Since the y-value is double the x-value, we write the point as \((x, 2x)\). We then use the distance formula to find \(x\), which gives two possible points.

Exam Tip: Keep in mind that 'abscissa' is the x-coordinate and 'ordinate' is the y-coordinate. Swapping these definitions will lead to an incorrect set of coordinates.

Question 7. A point \(P(2, -1)\) is equidistant from the points \(A(a, 7)\) and \(B(-3, a)\). Find \(a\).
Answer:
We are given that point \(P(2, -1)\) is at an equal distance from \(A(a, 7)\) and \(B(-3, a)\).
This implies:
\( PA = PB \)

\( \implies PA^2 = PB^2 \)
Applying the distance formula to both sides:
\( (a - 2)^2 + (7 - (-1))^2 = (-3 - 2)^2 + (a - (-1))^2 \)
\( (a - 2)^2 + (8)^2 = (-5)^2 + (a + 1)^2 \)
Expanding both sides:
\( a^2 - 4a + 4 + 64 = 25 + a^2 + 2a + 1 \)
\( a^2 - 4a + 68 = a^2 + 2a + 26 \)
Subtracting \( a^2 \) from both sides:
\( -4a + 68 = 2a + 26 \)
Rearranging the terms:
\( 68 - 26 = 2a + 4a \)
\( 42 = 6a \)
\( a = 7 \)
In simple words: Since point P is the same distance from both A and B, we set their squared distances equal to each other. Expanding the terms and simplifying allows us to solve for a.

Exam Tip: Squaring both sides of the distance equation is a useful step to eliminate the radical sign and simplify the algebraic calculations.

Question 8. What point on the x-axis is equidistant from the points \((7, 6)\) and \((-3, 4)\)?
Answer:
Let the required point on the x-axis be represented by \(P(x, 0)\).
The two given points are \(A(7, 6)\) and \(B(-3, 4)\).
Since \(P\) is equidistant from \(A\) and \(B\), we write:
\( PA = PB \)

\( \implies PA^2 = PB^2 \)
Using the distance formula:
\( (x - 7)^2 + (0 - 6)^2 = (x - (-3))^2 + (0 - 4)^2 \)
\( (x - 7)^2 + (-6)^2 = (x + 3)^2 + (-4)^2 \)
Expanding the terms:
\( x^2 - 14x + 49 + 36 = x^2 + 6x + 9 + 16 \)
\( x^2 - 14x + 85 = x^2 + 6x + 25 \)
Simplifying the equation by canceling \(x^2\) on both sides:
\( -14x + 85 = 6x + 25 \)
\( 85 - 25 = 6x + 14x \)
\( 60 = 20x \)
\( x = 3 \)
Thus, the coordinates of the required point on the x-axis are \((3, 0)\).
In simple words: We assume the point on the x-axis is \((x, 0)\). We equate its squared distance to both given points, cancel out the quadratic terms, and solve for x.

Exam Tip: Be careful to keep the y-coordinate as 0 since the point lies on the x-axis. This reduces the number of variables to one.

Question 9. Find a point on the y-axis which is equidistant from the points \((5, 2)\) and \((-4, 3)\).
Answer:
Let the coordinates of the point on the y-axis be \(P(0, y)\).
The given points are \(A(5, 2)\) and \(B(-4, 3)\).
Since \(P\) is equidistant from both \(A\) and \(B\), we have:
\( PA = PB \)

\( \implies PA^2 = PB^2 \)
Applying the distance formula:
\( (0 - 5)^2 + (y - 2)^2 = (0 - (-4))^2 + (y - 3)^2 \)
\( (-5)^2 + (y - 2)^2 = (4)^2 + (y - 3)^2 \)
Expanding both sides:
\( 25 + y^2 - 4y + 4 = 16 + y^2 - 6y + 9 \)
\( y^2 - 4y + 29 = y^2 - 6y + 25 \)
Subtracting \( y^2 \) from both sides:
\( -4y + 29 = -6y + 25 \)
Rearranging the terms:
\( 6y - 4y = 25 - 29 \)
\( 2y = -4 \)
\( y = -2 \)
Therefore, the coordinates of the required point are \((0, -2)\).
In simple words: We write the point on the y-axis as \((0, y)\). By setting the distance squared to both points equal, we find the value of y.

Exam Tip: Keep track of negative signs when moving terms across the equals sign. A small arithmetic error can change the final coordinates.

Question 10. 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.
Answer:
(i) Since point \( P \) is situated on the x-axis, its y-coordinate (ordinate) must be \( 0 \).
(ii) Since point \( Q \) is situated on the y-axis, its x-coordinate (abscissa) must be \( 0 \).
(iii) Given the abscissa of \( P \) is \(-12\) and the ordinate of \( Q \) is \(-16\), their coordinates are \( P(-12, 0) \) and \( Q(0, -16) \).
The length of the line segment \( PQ \) is calculated as:
\( PQ = \sqrt{(-12 - 0)^2 + (0 - (-16))^2} \)
\( PQ = \sqrt{(-12)^2 + (16)^2} \)
\( PQ = \sqrt{144 + 256} \)
\( PQ = \sqrt{400} = 20 \) units
In simple words: Any point on the x-axis has a y-value of 0, and any point on the y-axis has an x-value of 0. Using these rules, we find the coordinates of both points and calculate the distance between them.

Exam Tip: Clearly label each part of the question (i, ii, and iii) in your answers, as marks are allocated separately for each sub-part.

Question 11. Show that the points \(P(0, 5)\), \(Q(5, 10)\) and \(R(6, 3)\) are the vertices of an isosceles triangle.
Answer:
Let the coordinates of the vertices be \( P(0, 5) \), \( Q(5, 10) \), and \( R(6, 3) \).
Using the distance formula, we calculate the lengths of the sides of the triangle:
Side \( PQ \):
\( PQ = \sqrt{(5 - 0)^2 + (10 - 5)^2} \)
\( PQ = \sqrt{5^2 + 5^2} \)
\( PQ = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2} \) units

Side \( QR \):
\( QR = \sqrt{(6 - 5)^2 + (3 - 10)^2} \)
\( QR = \sqrt{1^2 + (-7)^2} \)
\( QR = \sqrt{1 + 49} = \sqrt{50} = 5\sqrt{2} \) units

Side \( RP \):
\( RP = \sqrt{(0 - 6)^2 + (5 - 3)^2} \)
\( RP = \sqrt{(-6)^2 + 2^2} \)
\( RP = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10} \) units

Since \( PQ = QR \), the triangle has two sides of equal length.
Thus, \(\Delta PQR\) is an isosceles triangle.
In simple words: An isosceles triangle has two sides that are equal in length. We calculate all three side lengths, and since two of them are the same, the triangle is isosceles.

Exam Tip: To prove a triangle is isosceles, show that exactly two sides are equal in length. Always simplify your square roots to their lowest terms.

Question 12. Prove that the points \(P(0, -4)\), \(Q(6, 2)\), \(R(3, 5)\) and \(S(-3, -1)\) are the vertices of a rectangle PQRS.
Answer:
We determine the lengths of the sides and diagonals of the quadrilateral \( PQRS \) using the distance formula:
Side \( PQ \):
\( PQ = \sqrt{(6 - 0)^2 + (2 - (-4))^2} \)
\( PQ = \sqrt{6^2 + 6^2} \)
\( PQ = \sqrt{36 + 36} = 6\sqrt{2} \) units

Side \( QR \):
\( QR = \sqrt{(6 - 3)^2 + (2 - 5)^2} \)
\( QR = \sqrt{3^2 + (-3)^2} \)
\( QR = \sqrt{9 + 9} = 3\sqrt{2} \) units

Side \( RS \):
\( RS = \sqrt{(3 - (-3))^2 + (5 - (-1))^2} \)
\( RS = \sqrt{6^2 + 6^2} \)
\( RS = \sqrt{36 + 36} = 6\sqrt{2} \) units

Side \( PS \):
\( PS = \sqrt{(-3 - 0)^2 + (-1 - (-4))^2} \)
\( PS = \sqrt{(-3)^2 + 3^2} \)
\( PS = \sqrt{9 + 9} = 3\sqrt{2} \) units

Diagonal \( PR \):
\( PR = \sqrt{(3 - 0)^2 + (5 - (-4))^2} \)
\( PR = \sqrt{3^2 + 9^2} \)
\( PR = \sqrt{9 + 81} = \sqrt{90} = 3\sqrt{10} \) units

Diagonal \( QS \):
\( QS = \sqrt{(6 - (-3))^2 + (2 - (-1))^2} \)
\( QS = \sqrt{9^2 + 3^2} \)
\( QS = \sqrt{81 + 9} = \sqrt{90} = 3\sqrt{10} \) units

Here, we observe that the opposite sides are equal:
\( PQ = RS \) and \( QR = PS \)
Also, the diagonals are equal in length:
\( PR = QS \)
Since both conditions are satisfied, \( PQRS \) is a rectangle.
In simple words: To show a shape is a rectangle, we prove that its opposite sides are equal and its diagonal lines are also equal in length. Since all these distances match, the points form a rectangle.

Exam Tip: Simply proving opposite sides are equal only shows the figure is a parallelogram. You must also prove that the diagonals are equal to confirm it is a rectangle.

Question 13. 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.
Answer:
We calculate the side lengths of triangle \( ABC \) using the distance formula:
Side \( AB \):
\( AB = \sqrt{(-3 - 1)^2 + (0 - (-3))^2} \)
\( AB = \sqrt{(-4)^2 + 3^2} \)
\( AB = \sqrt{16 + 9} = \sqrt{25} = 5 \) units

Side \( BC \):
\( BC = \sqrt{(4 - (-3))^2 + (1 - 0)^2} \)
\( BC = \sqrt{7^2 + 1^2} \)
\( BC = \sqrt{49 + 1} = \sqrt{50} = 5\sqrt{2} \) units

Side \( CA \):
\( CA = \sqrt{(1 - 4)^2 + (-3 - 1)^2} \)
\( CA = \sqrt{(-3)^2 + (-4)^2} \)
\( CA = \sqrt{9 + 16} = \sqrt{25} = 5 \) units

We see that:
\( AB = CA = 5 \) units
This indicates that \(\Delta ABC\) is an isosceles triangle.

Now, let us verify the Pythagoras theorem:
\( AB^2 + CA^2 = 5^2 + 5^2 = 25 + 25 = 50 \)
\( BC^2 = (5\sqrt{2})^2 = 50 \)
Since \( AB^2 + CA^2 = BC^2 \), the triangle is right-angled at vertex \( A \).
Thus, \(\Delta ABC\) is an isosceles right-angled triangle.

To find the area:
\(\text{Area of } \Delta ABC = \frac{1}{2} \times AB \times CA \)
\(\text{Area} = \frac{1}{2} \times 5 \times 5 \)
\(\text{Area} = 12.5 \text{ sq. units}\)
In simple words: First, we find the three side lengths. Since two sides are equal, it is isosceles. Since the square of the longest side equals the sum of the squares of the other two sides, it is right-angled. We then find the area using half of base times height.

Exam Tip: In a right-angled triangle, the two shorter sides are the base and height. Never use the hypotenuse (the longest side) when calculating the area.

Question 14. Show that the points \(A(5, 6)\), \(B(1, 5)\), \(C(2, 1)\) and \(D(6, 2)\) are the vertices of a square ABCD.
Answer:
We calculate the side lengths and diagonals of the quadrilateral \( ABCD \):
Side \( AB \):
\( AB = \sqrt{(1 - 5)^2 + (5 - 6)^2} \)
\( AB = \sqrt{(-4)^2 + (-1)^2} = \sqrt{16 + 1} = \sqrt{17} \) units

Side \( BC \):
\( BC = \sqrt{(2 - 1)^2 + (1 - 5)^2} \)
\( BC = \sqrt{1^2 + (-4)^2} = \sqrt{1 + 16} = \sqrt{17} \) units

Side \( CD \):
\( CD = \sqrt{(6 - 2)^2 + (2 - 1)^2} \)
\( CD = \sqrt{4^2 + 1^2} = \sqrt{16 + 1} = \sqrt{17} \) units

Side \( DA \):
\( DA = \sqrt{(5 - 6)^2 + (6 - 2)^2} \)
\( DA = \sqrt{(-1)^2 + 4^2} = \sqrt{1 + 16} = \sqrt{17} \) units

Now we compute the lengths of the diagonals:
Diagonal \( AC \):
\( AC = \sqrt{(2 - 5)^2 + (1 - 6)^2} \)
\( AC = \sqrt{(-3)^2 + (-5)^2} = \sqrt{9 + 25} = \sqrt{34} \) units

Diagonal \( BD \):
\( BD = \sqrt{(6 - 1)^2 + (2 - 5)^2} \)
\( BD = \sqrt{5^2 + (-3)^2} = \sqrt{25 + 9} = \sqrt{34} \) units

Since all four sides are equal:
\( AB = BC = CD = DA = \sqrt{17} \)
And both diagonals are equal:
\( AC = BD = \sqrt{34} \)
This proves that \( ABCD \) is a square.
In simple words: A square has four equal sides and two equal diagonals. We compute these six distances, and since all sides match and both diagonals are equal, the shape is a square.

Exam Tip: To prove a figure is a square, it is not enough to show that all four sides are equal (which only proves it is a rhombus). You must also show the diagonals are equal.

Question 15. Show that \((-3, 2)\), \((-5, -5)\), \((2, -3)\) and \((4, 4)\) are the vertices of a rhombus.
Answer:
Let the given points be labeled as \( A(-3, 2) \), \( B(-5, -5) \), \( C(2, -3) \), and \( D(4, 4) \).
Let us find the lengths of all four sides:
Side \( AB \):
\( AB = \sqrt{(-5 - (-3))^2 + (-5 - 2)^2} \)
\( AB = \sqrt{(-2)^2 + (-7)^2} = \sqrt{4 + 49} = \sqrt{53} \) units

Side \( BC \):
\( BC = \sqrt{(2 - (-5))^2 + (-3 - (-5))^2} \)
\( BC = \sqrt{7^2 + 2^2} = \sqrt{49 + 4} = \sqrt{53} \) units

Side \( CD \):
\( CD = \sqrt{(4 - 2)^2 + (4 - (-3))^2} \)
\( CD = \sqrt{2^2 + 7^2} = \sqrt{4 + 49} = \sqrt{53} \) units

Side \( DA \):
\( DA = \sqrt{(-3 - 4)^2 + (2 - 4)^2} \)
\( DA = \sqrt{(-7)^2 + (-2)^2} = \sqrt{49 + 4} = \sqrt{53} \) units

Now we find the lengths of the diagonals:
Diagonal \( AC \):
\( AC = \sqrt{(2 - (-3))^2 + (-3 - 2)^2} \)
\( AC = \sqrt{5^2 + (-5)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2} \) units

Diagonal \( BD \):
\( BD = \sqrt{(4 - (-5))^2 + (4 - (-5))^2} \)
\( BD = \sqrt{9^2 + 9^2} = \sqrt{81 + 81} = \sqrt{162} = 9\sqrt{2} \) units

We find that:
\( AB = BC = CD = DA = \sqrt{53} \)
But the diagonals are not equal:
\( AC \neq BD \)
Therefore, the vertices form a rhombus.
In simple words: A rhombus has four equal sides, but its diagonal lines are different lengths. Since our side lengths match but the diagonals do not, the shape is a rhombus.

Exam Tip: Be precise in showing that the diagonals are unequal. If they were equal, the figure would be classified as a square instead.

Question 16. 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 \(AB = CD\).
Answer:
We are given that \( AB = CD \).
Squaring both sides:
\( AB^2 = CD^2 \)
Using the distance formula:
\( (-6 - (-3))^2 + (a - (-2))^2 = (0 - (-3))^2 + (-1 - (-4))^2 \)
\( (-3)^2 + (a + 2)^2 = (3)^2 + (3)^2 \)
\( 9 + a^2 + 4a + 4 = 9 + 9 \)
\( a^2 + 4a + 13 = 18 \)
\( a^2 + 4a - 5 = 0 \)
Factoring the quadratic equation:
\( a^2 + 5a - a - 5 = 0 \)
\( a(a + 5) - 1(a + 5) = 0 \)
\( (a - 1)(a + 5) = 0 \)
This gives:
\( a = 1 \) or \( a = -5 \)
Since the problem specifies that \( a \) must be a negative number, we select:
\( a = -5 \)
In simple words: We set the squared distance of side AB equal to CD. Solving the resulting quadratic equation gives two answers, and we choose the negative one as specified.

Exam Tip: Always check the constraints given in the question. Here, 'a is negative' is the key constraint that lets us discard \( a = 1 \).

Question 17. The vertices of a triangle are \((5, 1)\), \((11, 1)\) and \((11, 9)\). Find the coordinates of the circumcentre of the triangle.
Answer:
Let the circumcentre of the triangle be denoted as \( P(x, y) \).
The circumcentre is equidistant from all three vertices \( A(5, 1) \), \( B(11, 1) \), and \( C(11, 9) \).
Therefore:
\( PA = PB \)

\( \implies PA^2 = PB^2 \)
Applying the distance formula:
\( (x - 5)^2 + (y - 1)^2 = (x - 11)^2 + (y - 1)^2 \)
Expanding the terms:
\( x^2 - 10x + 25 = x^2 - 22x + 121 \)
Simplifying by canceling \( x^2 \):
\( -10x + 22x = 121 - 25 \)
\( 12x = 96 \)
\( x = 8 \)

Now, we also have:
\( PA = PC \)

\( \implies PA^2 = PC^2 \)
Using the distance formula:
\( (x - 5)^2 + (y - 1)^2 = (x - 11)^2 + (y - 9)^2 \)
Expanding both sides:
\( x^2 - 10x + 25 + y^2 - 2y + 1 = x^2 - 22x + 121 + y^2 - 18y + 81 \)
\( -10x - 2y + 26 = -22x - 18y + 202 \)
Substitute \( x = 8 \) into this equation:
\( -10(8) - 2y + 26 = -22(8) - 18y + 202 \)
\( -80 - 2y + 26 = -176 - 18y + 202 \)
\( -54 - 2y = 26 - 18y \)
\( 18y - 2y = 26 + 54 \)
\( 16y = 80 \)
\( y = 5 \)
Thus, the coordinates of the circumcentre are \((8, 5)\).
In simple words: The circumcentre is a point that is the exact same distance from all three corners of the triangle. By setting up equations for these distances, we can solve for x and y.

Exam Tip: Since the y-coordinates of A and B are the same, \( PA^2 = PB^2 \) simplifies directly to find x, which makes the overall calculation much quicker.

Question 18. Given \(A(3, 1)\) and \(B(0, y - 1)\). Find \(y\) if \(AB = 5\).
Answer:
We are given \( AB = 5 \).
Squaring both sides:
\( AB^2 = 25 \)
Using the distance formula:
\( (0 - 3)^2 + ((y - 1) - 1)^2 = 25 \)
\( (-3)^2 + (y - 2)^2 = 25 \)
\( 9 + y^2 - 4y + 4 = 25 \)
\( y^2 - 4y + 13 - 25 = 0 \)
\( y^2 - 4y - 12 = 0 \)
Factoring this quadratic equation:
\( y^2 - 6y + 2y - 12 = 0 \)
\( y(y - 6) + 2(y - 6) = 0 \)
\( (y - 6)(y + 2) = 0 \)
This gives:
\( y = 6 \) or \( y = -2 \)
In simple words: We use the distance formula with the coordinates of A and B, set the square of the distance to 25, and solve the quadratic equation to find the values of y.

Exam Tip: Take care when subtracting the y-coordinates: \((y - 1) - 1\) simplifies to \(y - 2\). A sign mistake here will lead to incorrect factoring.

Question 19. Given \(A(x + 2, -2)\) and \(B(11, 6)\). Find \(x\) if \(AB = 17\).
Answer:
We are given \( AB = 17 \).
Squaring both sides:
\( AB^2 = 289 \)
Applying the distance formula:
\( (11 - (x + 2))^2 + (6 - (-2))^2 = 289 \)
\( (11 - x - 2)^2 + (6 + 2)^2 = 289 \)
\( (9 - x)^2 + 8^2 = 289 \)
\( 81 - 18x + x^2 + 64 = 289 \)
\( x^2 - 18x + 145 - 289 = 0 \)
\( x^2 - 18x - 144 = 0 \)
Factoring the quadratic equation:
\( x^2 - 24x + 6x - 144 = 0 \)
\( x(x - 24) + 6(x - 24) = 0 \)
\( (x - 24)(x + 6) = 0 \)
This results in:
\( x = 24 \) or \( x = -6 \)
In simple words: We write out the distance between the two points, square it to get rid of the root, and solve the quadratic equation to find the two possible values of x.

Exam Tip: Make sure to apply the negative sign to both terms when simplifying \(11 - (x + 2)\), which correctly yields \(9 - x\).

Question 20. Find the values of \(x\) if the distance between the points \(A(2x - 1, 3x + 1)\) and \(B(-3, -1)\) is equal to the radius of a circle whose diameter is 20 units.
Answer:
The distance between points \( A(2x - 1, 3x + 1) \) and \( B(-3, -1) \) is equal to the radius of the circle.
Since the diameter is 20 units, the radius is:
\(\text{Radius} = \frac{20}{2} = 10\) units
Therefore:
\( AB = 10 \)

\( \implies AB^2 = 100 \)
Using the distance formula:
\( (-3 - (2x - 1))^2 + (-1 - (3x + 1))^2 = 100 \)
\( (-3 - 2x + 1)^2 + (-1 - 3x - 1)^2 = 100 \)
\( (-2x - 2)^2 + (-3x - 2)^2 = 100 \)
Expanding the squared terms:
\( (4x^2 + 8x + 4) + (9x^2 + 12x + 4) = 100 \)
\( 13x^2 + 20x + 8 - 100 = 0 \)
\( 13x^2 + 20x - 92 = 0 \)
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\( x = \frac{-20 \pm \sqrt{20^2 - 4(13)(-92)}}{2(13)} \)
\( x = \frac{-20 \pm \sqrt{400 + 4784}}{26} \)
\( x = \frac{-20 \pm \sqrt{5184}}{26} \)
\( x = \frac{-20 \pm 72}{26} \)
Let us calculate the two possibilities:
- Taking the positive sign:
\( x = \frac{-20 + 72}{26} = \frac{52}{26} = 2 \)
- Taking the negative sign:
\( x = \frac{-20 - 72}{26} = \frac{-92}{26} = -\frac{46}{13} \)
Thus, \( x = 2 \) or \( x = -\frac{46}{13} \).
In simple words: Since the diameter is 20, the radius is 10. We set the distance between the two points to 10, expand the equation, and use the quadratic formula to solve for x.

Exam Tip: Keep your calculations neat when expanding \((-2x - 2)^2\). Since both terms are negative, squaring them results in all positive terms.

Question 21. Find the coordinates of point Q if the distance between the points \(P(2, -3)\) and \(Q(10, y)\) is 10 units.
Answer:
Let the coordinates of point \( Q \) be \((10, y)\).
The distance \( PQ \) is given as 10 units.
Therefore:
\( PQ = 10 \)

\( \implies PQ^2 = 100 \)
Using the distance formula:
\( (10 - 2)^2 + (y - (-3))^2 = 100 \)
\( (8)^2 + (y + 3)^2 = 100 \)
\( 64 + y^2 + 6y + 9 = 100 \)
\( y^2 + 6y + 73 - 100 = 0 \)
\( y^2 + 6y - 27 = 0 \)
Factorizing this quadratic equation:
\( y^2 + 9y - 3y - 27 = 0 \)
\( y(y + 9) - 3(y + 9) = 0 \)
\( (y + 9)(y - 3) = 0 \)
This yields:
\( y = -9 \) or \( y = 3 \)
Thus, the possible coordinates of point \( Q \) are \((10, -9)\) and \((10, 3)\).
In simple words: We use the distance formula to set up an equation equal to 10, square it, and then factor the quadratic equation to find the two possible values for y.

Exam Tip: Always provide both sets of coordinates in your final answer, as there are two distinct points on the line that satisfy the distance condition.

Question 22. The radius of a circle with center \(P(2, -7)\) is 13 units. \(AB\) is a chord of the circle and \(PT\) is drawn perpendicular to \(AB\), where \(T\) is \((-2, -4)\).
(i) Find the length of \(PT\) and \(AT\).
(ii) Find the length of the chord \(AB\).
Answer:
(i) We are given that the radius of the circle is 13 units, so:
\( PA = PB = 13 \) units
Using the distance formula, we find the length of the perpendicular \( PT \) from center \( P(2, -7) \) to point \( T(-2, -4) \):
\( PT = \sqrt{(-2 - 2)^2 + (-4 - (-7))^2} \)
\( = \sqrt{(-4)^2 + (3)^2} \)
\( = \sqrt{16 + 9} = \sqrt{25} = 5 \) units

In the right-angled triangle \(\Delta PAT\), we apply the Pythagoras theorem:
\( AT^2 = PA^2 - PT^2 \)
\( = 13^2 - 5^2 \)
\( = 169 - 25 = 144 \)
\( AT = 12 \) units

(ii) Since the perpendicular drawn from the center of a circle to any chord bisects that chord, we have:
\( AB = 2 \times AT \)
\( = 2 \times 12 = 24 \) units
In simple words: We first calculate the distance from the center to the chord. Using this distance and the radius, we use the Pythagoras theorem to find half of the chord's length, then double it to find the full length.

Exam Tip: Remember the circle property: a perpendicular from the center always bisects the chord. Clearly state this geometric reason in your solution.

Question 23. Find the distance between the points \((2, 2)\) and \((5, 4)\) correct to two decimal places.
Answer:
Using the distance formula, we calculate the distance between the two points:
\( PQ = \sqrt{(5 - 2)^2 + (4 - 2)^2} \)
\( = \sqrt{3^2 + 2^2} \)
\( = \sqrt{9 + 4} = \sqrt{13} \)
\( \approx 3.6055 \approx 3.61 \) units
In simple words: We plug the coordinates into the distance formula, find the square root of 13, and round the final answer to two decimal places.

Exam Tip: When asked for an answer to a specific number of decimal places, always carry out the square root calculation to one extra place before rounding.

Question 24. Point \(A\) is \((7, 3)\) and point \(B\) lies on the x-axis with an abscissa of 11. Find the distance \(AB\).
Answer:
Any point located on the x-axis has a y-coordinate of 0, meaning its coordinates are of the form \( (x, 0) \).
Since the abscissa (x-coordinate) of point \( B \) is given as 11, its coordinates are \((11, 0)\).
Now, using the distance formula between \( A(7, 3) \) and \( B(11, 0) \):
\( AB = \sqrt{(11 - 7)^2 + (0 - 3)^2} \)
\( = \sqrt{4^2 + (-3)^2} \)
\( = \sqrt{16 + 9} = \sqrt{25} = 5 \) units
In simple words: Since point B is on the x-axis with an x-value of 11, its coordinates are (11, 0). We then calculate the distance between (7, 3) and (11, 0).

Exam Tip: The term 'abscissa' refers strictly to the x-coordinate. Make sure to set the y-coordinate of any point on the x-axis to zero.

Question 25. Point \(A\) is \((5, -3)\) and point \(B\) lies on the y-axis with an ordinate of 9. Find the distance \(AB\).
Answer:
A point lying on the y-axis always has an x-coordinate of 0, so its coordinates are represented as \( (0, y) \).
Since the ordinate (y-coordinate) of point \( B \) is 9, the coordinates of \( B \) are \((0, 9)\).
Applying the distance formula to find the distance between \( A(5, -3) \) and \( B(0, 9) \):
\( AB = \sqrt{(0 - 5)^2 + (9 - (-3))^2} \)
\( = \sqrt{(-5)^2 + (12)^2} \)
\( = \sqrt{25 + 144} = \sqrt{169} = 13 \) units
In simple words: Since point B is on the y-axis with a y-value of 9, its coordinates are (0, 9). We then compute the distance between (5, -3) and (0, 9).

Exam Tip: 'Ordinate' refers to the y-coordinate. Always write the x-coordinate as 0 for any point on the y-axis.

Question 26. Find the coordinates of the points on the y-axis whose distances from the points \(A(6, 7)\) and \(B(4, -3)\) are in the ratio \(1 : 2\).
Answer:
Let the required point on the y-axis be \( P(0, y) \).
The distance from \( P \) to \( A(6, 7) \) is:
\( PA = \sqrt{(0 - 6)^2 + (y - 7)^2} \)
\( = \sqrt{36 + y^2 - 14y + 49} = \sqrt{y^2 - 14y + 85} \)

The distance from \( P \) to \( B(4, -3) \) is:
\( PB = \sqrt{(0 - 4)^2 + (y - (-3))^2} \)
\( = \sqrt{16 + (y + 3)^2} = \sqrt{16 + y^2 + 6y + 9} = \sqrt{y^2 + 6y + 25} \)

We are given that the ratio of these distances is \( 1 : 2 \):
\(\frac{PA}{PB} = \frac{1}{2}\)
Squaring both sides of the equation:
\(\frac{PA^2}{PB^2} = \frac{1}{4}\)

\( \implies 4 \times PA^2 = PB^2 \)
Substitute the expressions for \( PA^2 \) and \( PB^2 \):
\( 4(y^2 - 14y + 85) = y^2 + 6y + 25 \)
\( 4y^2 - 56y + 340 = y^2 + 6y + 25 \)
Rearranging terms to form a quadratic equation:
\( 3y^2 - 62y + 315 = 0 \)
Using the quadratic formula:
\( y = \frac{-(-62) \pm \sqrt{(-62)^2 - 4(3)(315)}}{2(3)} \)
\( = \frac{62 \pm \sqrt{3844 - 3780}}{6} \)
\( = \frac{62 \pm \sqrt{64}}{6} \)
\( = \frac{62 \pm 8}{6} \)
Let us calculate the two possible values of \( y \):
- For the positive sign:
\( y = \frac{62 + 8}{6} = \frac{70}{6} = \frac{35}{3} \)
- For the negative sign:
\( y = \frac{62 - 8}{6} = \frac{54}{6} = 9 \)
Thus, the required points on the y-axis are \((0, 9)\) and \(\left(0, \frac{35}{3}\right)\).
In simple words: We assume the point on the y-axis is \((0, y)\). Since the ratio of its distances to A and B is 1 to 2, we set up a squared ratio equation, simplify it into a quadratic equation, and solve for y.

Exam Tip: When dealing with ratios of distances, squaring both sides of the ratio equation is the easiest way to remove square roots and simplify the algebra.

Question 27. 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: \(5x^2 + 5y^2 - 34x + 70y + 58 = 0\).
Answer:
We are given the ratio of the distances:
\( PA : PB = 2 : 3 \)
This can be written as:
\(\frac{PA}{PB} = \frac{2}{3}\)
Squaring both sides:
\(\frac{PA^2}{PB^2} = \frac{4}{9}\)

\( \implies 9 \times PA^2 = 4 \times PB^2 \)
Applying the distance formula:
\( 9 \left[ (x - 1)^2 + (y - (-3))^2 \right] = 4 \left[ (x - (-2))^2 + (y - 2)^2 \right] \)
\( 9 \left[ (x - 1)^2 + (y + 3)^2 \right] = 4 \left[ (x + 2)^2 + (y - 2)^2 \right] \)
Expanding the squares inside the brackets:
\( 9 \left( x^2 - 2x + 1 + y^2 + 6y + 9 \right) = 4 \left( x^2 + 4x + 4 + y^2 - 4y + 4 \right) \)
\( 9 \left( x^2 - 2x + y^2 + 6y + 10 \right) = 4 \left( x^2 + 4x + y^2 - 4y + 8 \right) \)
Multiplying through by the coefficients:
\( 9x^2 - 18x + 9y^2 + 54y + 90 = 4x^2 + 16x + 4y^2 - 16y + 32 \)
Grouping all terms on the left-hand side:
\( (9x^2 - 4x^2) + (9y^2 - 4y^2) - 18x - 16x + 54y + 16y + 90 - 32 = 0 \)
\( 5x^2 + 5y^2 - 34x + 70y + 58 = 0 \)
Hence proved.
In simple words: We write the distance ratio as a fraction, square it to remove the roots, cross-multiply, expand the algebraic brackets, and group all terms to prove the required equation.

Exam Tip: Double-check the expansion of binomials like \((y + 3)^2\) and \((y - 2)^2\). Sign errors when distributing the 9 and 4 are very common.

Question 28. 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.
Answer:
We calculate the square of the lengths of sides \( AB \), \( BC \), and \( CA \) using the distance formula:
\( AB^2 = (a - 3)^2 + (-2 - 0)^2 \)
\( AB^2 = a^2 - 6a + 9 + 4 = a^2 - 6a + 13 \)

\( BC^2 = (4 - a)^2 + (-1 - (-2))^2 \)
\( BC^2 = 16 - 8a + a^2 + (1)^2 = a^2 - 8a + 17 \)

\( CA^2 = (3 - 4)^2 + (0 - (-1))^2 \)
\( CA^2 = (-1)^2 + (1)^2 = 1 + 1 = 2 \)

Since \(\Delta ABC\) is right-angled at vertex \( A \), we apply the Pythagoras theorem:
\( AB^2 + AC^2 = BC^2 \)
Substituting the expressions we found:
\( (a^2 - 6a + 13) + 2 = a^2 - 8a + 17 \)

\( \implies a^2 - 6a + 15 = a^2 - 8a + 17 \)
Subtracting \( a^2 \) from both sides:
\( -6a + 15 = -8a + 17 \)
Rearranging the terms:
\( 8a - 6a = 17 - 15 \)
\( 2a = 2 \)

\( \implies a = 1 \)
Therefore, the value of \( a \) is 1.
In simple words: Since the triangle is right-angled at A, we use the Pythagoras theorem where the square of the hypotenuse side BC equals the sum of the squares of the other two sides. Solving this equation gives us the value of a.

Exam Tip: Identify the correct hypotenuse. Since the right angle is at A, the hypotenuse is the opposite side, which is BC.