Table of Contents
Exercise 7.1 Solutions – Class X Mathematics
1. Find the distance between the following pair of points:
(i) (2, 3) and (4, 1)
Using the distance formula: \( \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2} \)
Here, \( (x_1, y_1) = (2, 3) \), \( (x_2, y_2) = (4, 1) \).
Distance = \( \sqrt{(4 – 2)^2 + (1 – 3)^2} = \sqrt{(2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \).
Using the distance formula: \( \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2} \)
Here, \( (x_1, y_1) = (2, 3) \), \( (x_2, y_2) = (4, 1) \).
Distance = \( \sqrt{(4 – 2)^2 + (1 – 3)^2} = \sqrt{(2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \).
(ii) (-5, 7) and (-1, 3)
Here, \( (x_1, y_1) = (-5, 7) \), \( (x_2, y_2) = (-1, 3) \).
Distance = \( \sqrt{(-1 – (-5))^2 + (3 – 7)^2} = \sqrt{(4)^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \).
Here, \( (x_1, y_1) = (-5, 7) \), \( (x_2, y_2) = (-1, 3) \).
Distance = \( \sqrt{(-1 – (-5))^2 + (3 – 7)^2} = \sqrt{(4)^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \).
(iii) (-2, -3) and (3, 2)
Here, \( (x_1, y_1) = (-2, -3) \), \( (x_2, y_2) = (3, 2) \).
Distance = \( \sqrt{(3 – (-2))^2 + (2 – (-3))^2} = \sqrt{(5)^2 + (5)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2} \).
Here, \( (x_1, y_1) = (-2, -3) \), \( (x_2, y_2) = (3, 2) \).
Distance = \( \sqrt{(3 – (-2))^2 + (2 – (-3))^2} = \sqrt{(5)^2 + (5)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2} \).
(iv) (a, b) and (-a, -b)
Here, \( (x_1, y_1) = (a, b) \), \( (x_2, y_2) = (-a, -b) \).
Distance = \( \sqrt{(-a – a)^2 + (-b – b)^2} = \sqrt{(-2a)^2 + (-2b)^2} = \sqrt{4a^2 + 4b^2} = 2\sqrt{a^2 + b^2} \).
Here, \( (x_1, y_1) = (a, b) \), \( (x_2, y_2) = (-a, -b) \).
Distance = \( \sqrt{(-a – a)^2 + (-b – b)^2} = \sqrt{(-2a)^2 + (-2b)^2} = \sqrt{4a^2 + 4b^2} = 2\sqrt{a^2 + b^2} \).
2. Find the distance between the points (0, 0) and (36, 15).
Here, \( (x_1, y_1) = (0, 0) \), \( (x_2, y_2) = (36, 15) \).
Distance = \( \sqrt{(36 – 0)^2 + (15 – 0)^2} = \sqrt{36^2 + 15^2} = \sqrt{1296 + 225} = \sqrt{1521} = 39 \).
Distance = \( \sqrt{(36 – 0)^2 + (15 – 0)^2} = \sqrt{36^2 + 15^2} = \sqrt{1296 + 225} = \sqrt{1521} = 39 \).
3. Verify whether the points (1, 5), (2, 3) and (-2, -1) are collinear or not.
Points are collinear if the area of the triangle formed by them is zero.
Using the formula: Area = \( \frac{1}{2} \left| x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2) \right| \)
Here, \( (x_1, y_1) = (1, 5) \), \( (x_2, y_2) = (2, 3) \), \( (x_3, y_3) = (-2, -1) \).
Area = \( \frac{1}{2} \left| 1(3 – (-1)) + 2(-1 – 5) + (-2)(5 – 3) \right| = \frac{1}{2} \left| 1(4) + 2(-6) + (-2)(2) \right| = \frac{1}{2} \left| 4 – 12 – 4 \right| = \frac{1}{2} \times 12 = 6 \).
Since the area is not zero, the points are not collinear.
Using the formula: Area = \( \frac{1}{2} \left| x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2) \right| \)
Here, \( (x_1, y_1) = (1, 5) \), \( (x_2, y_2) = (2, 3) \), \( (x_3, y_3) = (-2, -1) \).
Area = \( \frac{1}{2} \left| 1(3 – (-1)) + 2(-1 – 5) + (-2)(5 – 3) \right| = \frac{1}{2} \left| 1(4) + 2(-6) + (-2)(2) \right| = \frac{1}{2} \left| 4 – 12 – 4 \right| = \frac{1}{2} \times 12 = 6 \).
Since the area is not zero, the points are not collinear.
4. Check whether (5, -2), (6, 4) and (7, -2) are the vertices of an isosceles triangle.
Calculate the lengths of the sides using the distance formula.
\( A(5, -2) \), \( B(6, 4) \), \( C(7, -2) \).
\( AB = \sqrt{(6 – 5)^2 + (4 – (-2))^2} = \sqrt{1^2 + 6^2} = \sqrt{37} \).
\( BC = \sqrt{(7 – 6)^2 + (-2 – 4)^2} = \sqrt{1^2 + (-6)^2} = \sqrt{37} \).
\( CA = \sqrt{(7 – 5)^2 + (-2 – (-2))^2} = \sqrt{2^2 + 0^2} = 2 \).
Since \( AB = BC \), the triangle is isosceles.
\( A(5, -2) \), \( B(6, 4) \), \( C(7, -2) \).
\( AB = \sqrt{(6 – 5)^2 + (4 – (-2))^2} = \sqrt{1^2 + 6^2} = \sqrt{37} \).
\( BC = \sqrt{(7 – 6)^2 + (-2 – 4)^2} = \sqrt{1^2 + (-6)^2} = \sqrt{37} \).
\( CA = \sqrt{(7 – 5)^2 + (-2 – (-2))^2} = \sqrt{2^2 + 0^2} = 2 \).
Since \( AB = BC \), the triangle is isosceles.
5. In a classroom, 4 friends are seated at the points A, B, C and D as shown in Figure. Jarina and Phani walk into the class and after observing for a few minutes Jarina asks Phani “Don’t you notice that ABCD is a square?” Phani disagrees. Using distance formula, decide who is correct why?
From the figure: \( A(4, 5) \), \( B(8, 7) \), \( C(8, 5) \), \( D(4, 3) \).
A square has all sides equal and diagonals equal.
Sides: \( AB = \sqrt{(8 – 4)^2 + (7 – 5)^2} = \sqrt{16 + 4} = \sqrt{20} \),
\( BC = \sqrt{(8 – 8)^2 + (7 – 5)^2} = \sqrt{0 + 4} = 2 \),
\( CD = \sqrt{(8 – 4)^2 + (5 – 3)^2} = \sqrt{16 + 4} = \sqrt{20} \),
\( DA = \sqrt{(4 – 4)^2 + (5 – 3)^2} = \sqrt{0 + 4} = 2 \).
Diagonals: \( AC = \sqrt{(8 – 4)^2 + (5 – 5)^2} = 4 \),
\( BD = \sqrt{(8 – 4)^2 + (7 – 3)^2} = \sqrt{16 + 16} = \sqrt{32} \).
Sides \( AB = CD \) and \( BC = DA \), but diagonals are not equal, so ABCD is not a square. Phani is correct.
A square has all sides equal and diagonals equal.
Sides: \( AB = \sqrt{(8 – 4)^2 + (7 – 5)^2} = \sqrt{16 + 4} = \sqrt{20} \),
\( BC = \sqrt{(8 – 8)^2 + (7 – 5)^2} = \sqrt{0 + 4} = 2 \),
\( CD = \sqrt{(8 – 4)^2 + (5 – 3)^2} = \sqrt{16 + 4} = \sqrt{20} \),
\( DA = \sqrt{(4 – 4)^2 + (5 – 3)^2} = \sqrt{0 + 4} = 2 \).
Diagonals: \( AC = \sqrt{(8 – 4)^2 + (5 – 5)^2} = 4 \),
\( BD = \sqrt{(8 – 4)^2 + (7 – 3)^2} = \sqrt{16 + 16} = \sqrt{32} \).
Sides \( AB = CD \) and \( BC = DA \), but diagonals are not equal, so ABCD is not a square. Phani is correct.
6. Show that the following points form an equilateral triangle A(a, 0), B(-a, 0), C(0, a\sqrt{3}).
All sides of an equilateral triangle are equal.
\( AB = \sqrt{(-a – a)^2 + (0 – 0)^2} = \sqrt{(-2a)^2} = 2a \),
\( BC = \sqrt{(0 – (-a))^2 + (a\sqrt{3} – 0)^2} = \sqrt{a^2 + (a\sqrt{3})^2} = \sqrt{a^2 + 3a^2} = \sqrt{4a^2} = 2a \),
\( CA = \sqrt{(0 – a)^2 + (a\sqrt{3} – 0)^2} = \sqrt{a^2 + 3a^2} = 2a \).
Since \( AB = BC = CA \), the triangle is equilateral.
\( AB = \sqrt{(-a – a)^2 + (0 – 0)^2} = \sqrt{(-2a)^2} = 2a \),
\( BC = \sqrt{(0 – (-a))^2 + (a\sqrt{3} – 0)^2} = \sqrt{a^2 + (a\sqrt{3})^2} = \sqrt{a^2 + 3a^2} = \sqrt{4a^2} = 2a \),
\( CA = \sqrt{(0 – a)^2 + (a\sqrt{3} – 0)^2} = \sqrt{a^2 + 3a^2} = 2a \).
Since \( AB = BC = CA \), the triangle is equilateral.
7. Prove that the points (-7, -3), (5, 10), (15, 8) and (3, -5) taken in order are the corners of a parallelogram.
A parallelogram has opposite sides equal and parallel.
\( A(-7, -3) \), \( B(5, 10) \), \( C(15, 8) \), \( D(3, -5) \).
\( AB = \sqrt{(5 – (-7))^2 + (10 – (-3))^2} = \sqrt{12^2 + 13^2} = \sqrt{313} \),
\( CD = \sqrt{(15 – 3)^2 + (8 – (-5))^2} = \sqrt{12^2 + 13^2} = \sqrt{313} \),
\( BC = \sqrt{(15 – 5)^2 + (8 – 10)^2} = \sqrt{10^2 + (-2)^2} = \sqrt{104} \),
\( DA = \sqrt{(3 – (-7))^2 + (-5 – (-3))^2} = \sqrt{10^2 + (-2)^2} = \sqrt{104} \).
Since \( AB = CD \) and \( BC = DA \), the points form a parallelogram.
\( A(-7, -3) \), \( B(5, 10) \), \( C(15, 8) \), \( D(3, -5) \).
\( AB = \sqrt{(5 – (-7))^2 + (10 – (-3))^2} = \sqrt{12^2 + 13^2} = \sqrt{313} \),
\( CD = \sqrt{(15 – 3)^2 + (8 – (-5))^2} = \sqrt{12^2 + 13^2} = \sqrt{313} \),
\( BC = \sqrt{(15 – 5)^2 + (8 – 10)^2} = \sqrt{10^2 + (-2)^2} = \sqrt{104} \),
\( DA = \sqrt{(3 – (-7))^2 + (-5 – (-3))^2} = \sqrt{10^2 + (-2)^2} = \sqrt{104} \).
Since \( AB = CD \) and \( BC = DA \), the points form a parallelogram.
8. Show that the points (-4, -7), (-1, 2), (8, 5) and (5, -4) taken in order are the vertices of a rhombus. Find its area.
A rhombus has all sides equal. Area = \( \frac{1}{2} \times \text{product of diagonals} \).
\( A(-4, -7) \), \( B(-1, 2) \), \( C(8, 5) \), \( D(5, -4) \).
Sides: \( AB = \sqrt{(-1 – (-4))^2 + (2 – (-7))^2} = \sqrt{3^2 + 9^2} = \sqrt{90} \),
\( BC = \sqrt{(8 – (-1))^2 + (5 – 2)^2} = \sqrt{9^2 + 3^2} = \sqrt{90} \),
\( CD = \sqrt{(8 – 5)^2 + (5 – (-4))^2} = \sqrt{3^2 + 9^2} = \sqrt{90} \),
\( DA = \sqrt{(5 – (-4))^2 + (-4 – (-7))^2} = \sqrt{9^2 + 3^2} = \sqrt{90} \).
All sides are equal, so it’s a rhombus.
Diagonals: \( AC = \sqrt{(8 – (-4))^2 + (5 – (-7))^2} = \sqrt{12^2 + 12^2} = 12\sqrt{2} \),
\( BD = \sqrt{(5 – (-1))^2 + (-4 – 2)^2} = \sqrt{6^2 + (-6)^2} = 6\sqrt{2} \).
Area = \( \frac{1}{2} \times (12\sqrt{2}) \times (6\sqrt{2}) = \frac{1}{2} \times 72 \times 2 = 72 \) square units.
\( A(-4, -7) \), \( B(-1, 2) \), \( C(8, 5) \), \( D(5, -4) \).
Sides: \( AB = \sqrt{(-1 – (-4))^2 + (2 – (-7))^2} = \sqrt{3^2 + 9^2} = \sqrt{90} \),
\( BC = \sqrt{(8 – (-1))^2 + (5 – 2)^2} = \sqrt{9^2 + 3^2} = \sqrt{90} \),
\( CD = \sqrt{(8 – 5)^2 + (5 – (-4))^2} = \sqrt{3^2 + 9^2} = \sqrt{90} \),
\( DA = \sqrt{(5 – (-4))^2 + (-4 – (-7))^2} = \sqrt{9^2 + 3^2} = \sqrt{90} \).
All sides are equal, so it’s a rhombus.
Diagonals: \( AC = \sqrt{(8 – (-4))^2 + (5 – (-7))^2} = \sqrt{12^2 + 12^2} = 12\sqrt{2} \),
\( BD = \sqrt{(5 – (-1))^2 + (-4 – 2)^2} = \sqrt{6^2 + (-6)^2} = 6\sqrt{2} \).
Area = \( \frac{1}{2} \times (12\sqrt{2}) \times (6\sqrt{2}) = \frac{1}{2} \times 72 \times 2 = 72 \) square units.
9. 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)
\( A(-1, -2) \), \( B(1, 0) \), \( C(-1, 2) \), \( D(-3, 0) \).
Sides: \( AB = \sqrt{(1 – (-1))^2 + (0 – (-2))^2} = \sqrt{4 + 4} = 2\sqrt{2} \),
\( BC = \sqrt{(-1 – 1)^2 + (2 – 0)^2} = \sqrt{4 + 4} = 2\sqrt{2} \),
\( CD = \sqrt{(-1 – (-3))^2 + (2 – 0)^2} = \sqrt{4 + 4} = 2\sqrt{2} \),
\( DA = \sqrt{(-3 – (-1))^2 + (0 – (-2))^2} = \sqrt{4 + 4} = 2\sqrt{2} \).
All sides are equal, so it’s a rhombus.
Diagonals: \( AC = \sqrt{(-1 – (-1))^2 + (2 – (-2))^2} = 4 \), \( BD = \sqrt{(-3 – 1)^2 + (0 – 0)^2} = 4 \).
Diagonals are equal, so it’s a square.
\( A(-1, -2) \), \( B(1, 0) \), \( C(-1, 2) \), \( D(-3, 0) \).
Sides: \( AB = \sqrt{(1 – (-1))^2 + (0 – (-2))^2} = \sqrt{4 + 4} = 2\sqrt{2} \),
\( BC = \sqrt{(-1 – 1)^2 + (2 – 0)^2} = \sqrt{4 + 4} = 2\sqrt{2} \),
\( CD = \sqrt{(-1 – (-3))^2 + (2 – 0)^2} = \sqrt{4 + 4} = 2\sqrt{2} \),
\( DA = \sqrt{(-3 – (-1))^2 + (0 – (-2))^2} = \sqrt{4 + 4} = 2\sqrt{2} \).
All sides are equal, so it’s a rhombus.
Diagonals: \( AC = \sqrt{(-1 – (-1))^2 + (2 – (-2))^2} = 4 \), \( BD = \sqrt{(-3 – 1)^2 + (0 – 0)^2} = 4 \).
Diagonals are equal, so it’s a square.
(ii) (-3, 5), (3, 1), (1, -3), (-5, 1)
\( A(-3, 5) \), \( B(3, 1) \), \( C(1, -3) \), \( D(-5, 1) \).
Sides: \( AB = \sqrt{(3 – (-3))^2 + (1 – 5)^2} = \sqrt{36 + 16} = \sqrt{52} \),
\( BC = \sqrt{(1 – 3)^2 + (-3 – 1)^2} = \sqrt{4 + 16} = \sqrt{20} \),
\( CD = \sqrt{(-5 – 1)^2 + (1 – (-3))^2} = \sqrt{36 + 16} = \sqrt{52} \),
\( DA = \sqrt{(-5 – (-3))^2 + (1 – 5)^2} = \sqrt{4 + 16} = \sqrt{20} \).
Opposite sides are equal (\( AB = CD \), \( BC = DA \)), so it’s a parallelogram.
\( A(-3, 5) \), \( B(3, 1) \), \( C(1, -3) \), \( D(-5, 1) \).
Sides: \( AB = \sqrt{(3 – (-3))^2 + (1 – 5)^2} = \sqrt{36 + 16} = \sqrt{52} \),
\( BC = \sqrt{(1 – 3)^2 + (-3 – 1)^2} = \sqrt{4 + 16} = \sqrt{20} \),
\( CD = \sqrt{(-5 – 1)^2 + (1 – (-3))^2} = \sqrt{36 + 16} = \sqrt{52} \),
\( DA = \sqrt{(-5 – (-3))^2 + (1 – 5)^2} = \sqrt{4 + 16} = \sqrt{20} \).
Opposite sides are equal (\( AB = CD \), \( BC = DA \)), so it’s a parallelogram.
(iii) (4, 5), (7, 6), (4, 3), (1, 2)
\( A(4, 5) \), \( B(7, 6) \), \( C(4, 3) \), \( D(1, 2) \).
Sides: \( AB = \sqrt{(7 – 4)^2 + (6 – 5)^2} = \sqrt{9 + 1} = \sqrt{10} \),
\( BC = \sqrt{(4 – 7)^2 + (3 – 6)^2} = \sqrt{9 + 9} = 3\sqrt{2} \),
\( CD = \sqrt{(4 – 1)^2 + (3 – 2)^2} = \sqrt{9 + 1} = \sqrt{10} \),
\( DA = \sqrt{(1 – 4)^2 + (2 – 5)^2} = \sqrt{9 + 9} = 3\sqrt{2} \).
Opposite sides are equal, so it’s a parallelogram.
\( A(4, 5) \), \( B(7, 6) \), \( C(4, 3) \), \( D(1, 2) \).
Sides: \( AB = \sqrt{(7 – 4)^2 + (6 – 5)^2} = \sqrt{9 + 1} = \sqrt{10} \),
\( BC = \sqrt{(4 – 7)^2 + (3 – 6)^2} = \sqrt{9 + 9} = 3\sqrt{2} \),
\( CD = \sqrt{(4 – 1)^2 + (3 – 2)^2} = \sqrt{9 + 1} = \sqrt{10} \),
\( DA = \sqrt{(1 – 4)^2 + (2 – 5)^2} = \sqrt{9 + 9} = 3\sqrt{2} \).
Opposite sides are equal, so it’s a parallelogram.
10. Find the point on the X-axis which is equidistant from (2, -5) and (-2, 9).
Let the point on the X-axis be \( (x, 0) \).
Distance from \( (x, 0) \) to \( (2, -5) \): \( \sqrt{(x – 2)^2 + (0 – (-5))^2} = \sqrt{(x – 2)^2 + 25} \).
Distance from \( (x, 0) \) to \( (-2, 9) \): \( \sqrt{(x – (-2))^2 + (0 – 9)^2} = \sqrt{(x + 2)^2 + 81} \).
Since the distances are equal: \( \sqrt{(x – 2)^2 + 25} = \sqrt{(x + 2)^2 + 81} \).
Square both sides: \( (x – 2)^2 + 25 = (x + 2)^2 + 81 \),
\( x^2 – 4x + 4 + 25 = x^2 + 4x + 4 + 81 \),
\( -4x + 29 = 4x + 85 \),
\( -8x = 56 \), \( x = -7 \).
The point is \( (-7, 0) \).
Distance from \( (x, 0) \) to \( (2, -5) \): \( \sqrt{(x – 2)^2 + (0 – (-5))^2} = \sqrt{(x – 2)^2 + 25} \).
Distance from \( (x, 0) \) to \( (-2, 9) \): \( \sqrt{(x – (-2))^2 + (0 – 9)^2} = \sqrt{(x + 2)^2 + 81} \).
Since the distances are equal: \( \sqrt{(x – 2)^2 + 25} = \sqrt{(x + 2)^2 + 81} \).
Square both sides: \( (x – 2)^2 + 25 = (x + 2)^2 + 81 \),
\( x^2 – 4x + 4 + 25 = x^2 + 4x + 4 + 81 \),
\( -4x + 29 = 4x + 85 \),
\( -8x = 56 \), \( x = -7 \).
The point is \( (-7, 0) \).
11. If the distance between two points (x, 7) and (1, 15) is 10, find the value of x.
Distance = \( \sqrt{(x – 1)^2 + (7 – 15)^2} = 10 \).
\( \sqrt{(x – 1)^2 + (-8)^2} = 10 \),
\( (x – 1)^2 + 64 = 100 \),
\( (x – 1)^2 = 36 \),
\( x – 1 = \pm 6 \),
\( x = 7 \) or \( x = -5 \).
\( \sqrt{(x – 1)^2 + (-8)^2} = 10 \),
\( (x – 1)^2 + 64 = 100 \),
\( (x – 1)^2 = 36 \),
\( x – 1 = \pm 6 \),
\( x = 7 \) or \( x = -5 \).
12. Find the values of y for which the distance between the points P(2, -3) and Q(10, y) is 10 units.
Distance = \( \sqrt{(10 – 2)^2 + (y – (-3))^2} = 10 \).
\( \sqrt{8^2 + (y + 3)^2} = 10 \),
\( 64 + (y + 3)^2 = 100 \),
\( (y + 3)^2 = 36 \),
\( y + 3 = \pm 6 \),
\( y = 3 \) or \( y = -9 \).
\( \sqrt{8^2 + (y + 3)^2} = 10 \),
\( 64 + (y + 3)^2 = 100 \),
\( (y + 3)^2 = 36 \),
\( y + 3 = \pm 6 \),
\( y = 3 \) or \( y = -9 \).
13. Find the radius of the circle whose centre is (3, 2) and passes through (-5, 6).
Radius = distance between the center \( (3, 2) \) and the point \( (-5, 6) \).
\( \sqrt{(-5 – 3)^2 + (6 – 2)^2} = \sqrt{(-8)^2 + 4^2} = \sqrt{64 + 16} = \sqrt{80} = 4\sqrt{5} \).
The radius is \( 4\sqrt{5} \).
\( \sqrt{(-5 – 3)^2 + (6 – 2)^2} = \sqrt{(-8)^2 + 4^2} = \sqrt{64 + 16} = \sqrt{80} = 4\sqrt{5} \).
The radius is \( 4\sqrt{5} \).
14. Can you draw a triangle with vertices (1, 5), (5, 8) and (13, 14)? Give reason.
Check if the points are collinear using the area formula.
\( (x_1, y_1) = (1, 5) \), \( (x_2, y_2) = (5, 8) \), \( (x_3, y_3) = (13, 14) \).
Area = \( \frac{1}{2} \left| 1(8 – 14) + 5(14 – 5) + 13(5 – 8) \right| = \frac{1}{2} \left| 1(-6) + 5(9) + 13(-3) \right| = \frac{1}{2} \left| -6 + 45 – 39 \right| = 0 \).
Since the area is zero, the points are collinear, and a triangle cannot be formed.
\( (x_1, y_1) = (1, 5) \), \( (x_2, y_2) = (5, 8) \), \( (x_3, y_3) = (13, 14) \).
Area = \( \frac{1}{2} \left| 1(8 – 14) + 5(14 – 5) + 13(5 – 8) \right| = \frac{1}{2} \left| 1(-6) + 5(9) + 13(-3) \right| = \frac{1}{2} \left| -6 + 45 – 39 \right| = 0 \).
Since the area is zero, the points are collinear, and a triangle cannot be formed.
15. Find a relation between x and y such that the point (x, y) is equidistant from the points (-2, 8) and (-3, -5).
Distance from \( (x, y) \) to \( (-2, 8) \): \( \sqrt{(x – (-2))^2 + (y – 8)^2} \).
Distance from \( (x, y) \) to \( (-3, -5) \): \( \sqrt{(x – (-3))^2 + (y – (-5))^2} \).
\( \sqrt{(x + 2)^2 + (y – 8)^2} = \sqrt{(x + 3)^2 + (y + 5)^2} \).
Square both sides: \( (x + 2)^2 + (y – 8)^2 = (x + 3)^2 + (y + 5)^2 \),
\( x^2 + 4x + 4 + y^2 – 16y + 64 = x^2 + 6x + 9 + y^2 + 10y + 25 \),
\( 4x – 16y + 68 = 6x + 10y + 34 \),
\( -2x – 26y + 34 = 0 \),
\( x + 13y – 17 = 0 \).
Distance from \( (x, y) \) to \( (-3, -5) \): \( \sqrt{(x – (-3))^2 + (y – (-5))^2} \).
\( \sqrt{(x + 2)^2 + (y – 8)^2} = \sqrt{(x + 3)^2 + (y + 5)^2} \).
Square both sides: \( (x + 2)^2 + (y – 8)^2 = (x + 3)^2 + (y + 5)^2 \),
\( x^2 + 4x + 4 + y^2 – 16y + 64 = x^2 + 6x + 9 + y^2 + 10y + 25 \),
\( 4x – 16y + 68 = 6x + 10y + 34 \),
\( -2x – 26y + 34 = 0 \),
\( x + 13y – 17 = 0 \).
