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
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