Problems on Distance Between Two Points

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 




11 and 12 Grade Math 

From Problems on Distance Between Two Points to HOME PAGE




Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.



New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.

Share this page: What’s this?

Recent Articles

  1. Formation of Greatest and Smallest Numbers | Arranging the Numbers

    May 19, 24 03:36 PM

    Formation of Greatest and Smallest Numbers
    the greatest number is formed by arranging the given digits in descending order and the smallest number by arranging them in ascending order. The position of the digit at the extreme left of a number…

    Read More

  2. Formation of Numbers with the Given Digits |Making Numbers with Digits

    May 19, 24 03:19 PM

    In formation of numbers with the given digits we may say that a number is an arranged group of digits. Numbers may be formed with or without the repetition of digits.

    Read More

  3. Arranging Numbers | Ascending Order | Descending Order |Compare Digits

    May 19, 24 02:23 PM

    Arranging Numbers
    We know, while arranging numbers from the smallest number to the largest number, then the numbers are arranged in ascending order. Vice-versa while arranging numbers from the largest number to the sma…

    Read More

  4. Comparison of Numbers | Compare Numbers Rules | Examples of Comparison

    May 19, 24 01:26 PM

    Rules for Comparison of Numbers
    Rule I: We know that a number with more digits is always greater than the number with less number of digits. Rule II: When the two numbers have the same number of digits, we start comparing the digits…

    Read More

  5. Worksheets on Comparison of Numbers | Find the Greatest Number

    May 19, 24 10:42 AM

    Comparison of Two Numbers
    In worksheets on comparison of numbers students can practice the questions for fourth grade to compare numbers. This worksheet contains questions on numbers like to find the greatest number, arranging…

    Read More