Solving the problems on distance between two points with the help of the formula, in the below examples use the formula to find distance between two points.

**Worked-out problems on distance between two points:**

**1. Show that the points (3, 0), (6, 4) and (- 1, 3 ) are the vertices of a right-angled Isosceles triangle. Solution:** Let the given points be A(3, 0), B (6, 4) and C (-1, 3). Then we have,

AB² = (6 - 3)² + (4 - 0)² = 9 + 16 = 25;

BC² = (-1 - 6)² + (3 - 4 )² = 49 + 1= 50

and CA² = (3 + 1)² + (0 - 3)² = 16 + 9= 25.

From the above results we get,

AB² = CA² i.e., AB = CA,

which proves that the triangle ABC is isosceles.

Again, AB² + AC² = 25 + 25 = 50 = BC²

which shows that the triangle ABC is right-angled.

Therefore, the triangle formed by joining the given points is a right-angled isosceles triangle. *Proved*.

**2. If the three points (a, b), (a + k cos α, b + k sin α) and (a + k cos β, b + k sin β) are the vertices of an equilateral triangle, then which of the following is true and why ? **

**(i) | α - β| = π/4
(ii) |α - β| = π/2 (iii) |α - β| = π/6 (iv) |α - β| = π/3
Solution:**

Let the vertices of the triangle be A (a, b), B (a + k cos α, b + k sin α) and C (a + k cos β, b + k sin β).

Now, AB² = (a + k cos α - a)² + (b + k sin α - b)²

= k² cos² α + k² sin² α = k²;

Similarly, CA² = k² and

BC² = (a + k cos β - a - k cos α)² + (b + k sin β - b - k sin α)²

= k² (cos² β + cos² α - 2 cos α cos β + sin² β + sin² α - 2 sin α sin β)

= k² [cos² β + sin² β + cos² α + sin² α - 2(cos α cos β + sin α sin β)]

= k² [1 + 1 - 2 cos (α - β)]

= 2k² [1 - cos (α - β)]

Since ABC is an equilateral triangle, hence

AB² = BC²

or, k² = 2k² [1 - cos (α - β)]

or, 1/2 = 1 - cos(α - β) [since, k # 0]

or, cos (α - β) = 1/2 = cos π/3

Therefore, |α - β| = π/3 .

There for, condition (iv) is true.

**3. Find the point on the y-axis which is equidistant from the points (2, 3)and(-1, 2).
Solution:**

Let P(0, y) be the required point on the y-axis and the given points are A (2, 3) and B(- 1, 2). By question,

PA = PB = PA² = PB²

or, (2 - 0)² + (3 - y)² = (-1 - 0)² + (2 – y)²

or, 4 + 9 + y² - 6y = 1 + 4 + y² - 4y

or, - 6y + 4y = 1 - 9 or, - 2y = -8

or, y = 4.

Therefore, the required point on the y-axis is (0, 4).

**4. Find the circum-centre and circum-radius of the triangle whose vertices are (3, 4), (3, - 6) and (- 1, 2). **

Solution:

Let A(3, 4), B (3, - 6), C (- 1, 2) be the vertices of the triangle and P(x, y ) the required circum-centre and r the circum-radius. Then, we must have,

r² = PA² = (x - 3)² + (y - 4)² ……………………..(1)

r² = PB² = (x - 3)² + (y + 6)² ……………………….(2)

and r² = PC² = (x + 1)² + (y - 2)² ……………………….(3)

From (1) and (2) we get,

(x - 3)² + (y - 4)² = (x - 3)² + (y + 6)²

Or, y² - 8y + 16 = y² + 12y + 36

or, - 20y = 20 or, y = - 1

Again, from (2) and (3) we get,

(x - 3)² + (y + 6)² = (x + 1 )² + (y - 2)²

or, x² - 6x + 9 + 25 = x² + 2x + 1 + 9 [putting y = - 1]

or, - 8x = - 24

or, x = 3

Finally, putting x = 3 and y = - 1 in (1) we get,

r² = 0² + (-1 - 4)² = 25

Therefore, r = 5

Therefore, the co-ordinates of circum-centre are (3, - 1) and circum-radius = 5 units.

**5. Show that the four points (2, 5), (5, 9), (9, 12) and (6, 8) when joined in order, form a rhombus. Solution:**

Let the given points be A(2, 5), B (5, 9), C (9, 12) and D(6, 8). Now, AB² = (5 - 2)² + (9 - 5)² = 9 + 16 = 25

BC² = (9 - 5)² + (12 - 9)² = 16 + 9 = 25

CD² = (6 - 9)² (8 - 12)² = 9 + 16 = 25

DA² = (2 - 6)² + (5 - 8)² = 16 + 9 = 25

AC² = ( 9 - 2)² + (12 - 5)² = 49 + 49 = 98

and BD² = (6 - 5)² + (8 - 9)² = 1 + 1 = 2

From the above result we see that

AB = BC = CD = DA and AC ≠ BD.

That is the four sides of the quadrilateral ABCD are equal but diagonals AC and BD are not equal. Therefore, the quadrilateral ABCD is a rhombus. *Proved*.

The above worked-out problems on distance between two points are explained step-by-step with the help of the formula.

**●**** Co-ordinate Geometry**

**What is Co-ordinate Geometry?****Rectangular Cartesian Co-ordinates****Polar Co-ordinates****Relation between Cartesian and Polar Co-Ordinates****Distance between Two given Points****Distance between Two Points in Polar Co-ordinates****Division of Line Segment****: Internal & External****Area of the Triangle Formed by Three co-ordinate Points****Condition of Collinearity of Three Points****Medians of a Triangle are Concurrent****Apollonius' Theorem****Quadrilateral form a Parallelogram****Problems on Distance Between Two Points****Area of a Triangle Given 3 Points****Worksheet on Quadrants****Worksheet on Rectangular – Polar Conversion****Worksheet on Line-Segment Joining the Points****Worksheet on Distance Between Two Points****Worksheet on Distance Between the Polar Co-ordinates****Worksheet on Finding Mid-Point****Worksheet on Division of Line-Segment****Worksheet on Centroid of a Triangle****Worksheet on Area of Co-ordinate Triangle****Worksheet on Collinear Triangle****Worksheet on Area of Polygon****Worksheet on Cartesian Triangle**

**11 and 12 Grade Math**** ****From Problems on Distance Between Two Points to HOME PAGE**

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