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.
Workedout problems on distance between two points:
1. Show that the points (3, 0), (6, 4) and ( 1, 3 ) are the vertices of a rightangled 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 rightangled.
Therefore, the triangle formed by joining the given points is a rightangled 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 yaxis which is equidistant from the points (2, 3)and(1, 2).
Solution:
Let P(0, y) be the required point on the yaxis 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 yaxis is (0, 4).
4. Find the circumcentre and circumradius 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 circumcentre and r the circumradius. 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 coordinates of circumcentre are (3,  1) and circumradius = 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 workedout problems on distance between two points are explained stepbystep with the help of the formula.
● Coordinate Geometry
11 and 12 Grade Math
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