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Topics Covered in Distance and Section Formulae
- Distance Formula
- Distance Properties in some Geometrical Figures
- Conditions of Collinearity of Three Points
- Problems on Distance Formula
- Distance of a Point from the Origin
- Distance Formula in Geometry
- Section Formula
- Midpoint Formula
- Centroid of a Triangle
- Worksheet on Distance Formula
- Worksheet on Collinearity of Three Points
- Worksheet on Finding the Centroid of a Triangle
- Worksheet on Section Formula
Distance Formula
The distance formula is used to calculate the distance between the points on the two-dimensional and three-dimensional coordinate plane. You can find the distance between the points by substituting the distance formula.
d = √(x2 – x1)² + (y2 – y1)²
where,
d = distance between the points
(x1, y1) and (x2, y2) are the ordered pairs of the two coordinate planes.
Distance between three points formula:
The distance formula for two points is used to find the distance between two points A(x1, y1, z1) and B(x2, y2, z3).
d = √(x2 – x1)² + (y2 – y1)² + (z2 – z1)²
where,
d = distance between the points
(x1, y1, z1) and (x2, y2, z2) are the ordered pairs of the two coordinate planes.
Section Formula
The Section formula is used to find the ratio in which a line segment is divided by a point internally or externally. This formula is used to find the centroid, incenter, and excenters of a triangle.
Do Check: Co-ordinate Geometry
Distance and Section Formulae Solved Problems
Example 1.
Given a triangle ABC in which A = (4,-4) B = (0,5) and C(5,6) A point P lies on BC such that BP : PC = 3:2 find the length of line segment AP.
Solution:
Given BP : PC = 3 : 2
Using the section formula
The coordinates of the point p(x,y) divide the line segment joining the points A(x1,y1) and B(x2,y2) internally in the ratio m1: m2.
[m1x2 + m2x1/m2 + m1 , m1y2 + m2y1/m2 + m1]
m1 = 3
m2 = 2
x1 = 0
x2 = 5
y1 = 5
y2 = 10
[3×5 + 2×0/3 + 2, 3×10 + 2×5/3 + 2]
[15/2 , 40/5]
(3,8)
Using distance formula we have
AP = √(3 – 4)² + (8 + 4)²
√1 + 144
√145
= 12.04
Example 2.
A(10,0) and B(5,15) are two fixed points. Find the coordinates of the point P is AB such that 5PB = AB also find the coordinates of some other points Q in AB such that AB = 6AQ.
Solution:
Given that,
5PB = AB
AB/PB = 5/1
AB – PB/PB = 5 – 1/1
AB/PB = 4/1
Using section formula coordinates of P are
P(x,y) = P(4×10 + 1×5/4 + 1 , 4×15 + 1 × 0/4 + 1
(40 + 5/5 , 60/5)
(9,12)
Given AB = 6AQ
AQ/AB = 1/6
AQ/AB – AQ = 1/6 – 1
AQ/QB = 1/5
Using section formula coordinates of Q are
Q(x,y) = Q(1×10 + 5×5/1 + 5, 1×15 + 5×0/1 + 5
Q(x,y) = Q(10 + 25/6, 15/6)
Q(35/6, 15/6)
Example 3.
Coordinates of the points which Divide the join of (1,7) and (4,3) in the ratio of 2:3 is.
Solution:
We know that the coordinates of the point dividing (x1, y1) and (x2,y2) in the ratio m:n is given by
(m1x1 + m2xx1/m1 + m2 , m1y2 + m2y1/m1 + m2)
Given that
(x1,y1) = (1,7)
(x2,y2) = (4,3)
(m1, m2) = 2 : 3
Coordinates of point of intersection of line segment is
(2×4 + 3×1/2 + 3, 2×3 + 3×7/2 + 3)
(8 + 3/5, 6 + 21/5)
(11/5, 27/5)
Example 4.
Using the section formula, the points A(7,5) B(9,3) and C(13,1) are collinear.
Solution:
If three points are collinear, then one of the points divides the line segment joining the order points in the ratio r : 1.
If the P is between the line A and B and AP/PB = r.
Distance AB = √(x1 – x2)² + (y2 – y1)²
= √(9 – 7)² + (3 – 5)²
= √(7)² + (-2)²
= √49 + 4
= √53
= 7.28
Distance BC = √(13 – 9)² + (1 – 3)²
= √(4)² + (-2)²
= √16 + 4
= √20
To find r
r = AB/BC = 7.28/4.47
The line divides in the ratio of 7.28 : 4.47
A line divides internally in the ratio m : n the point P = (mx1 + nx1/m + n, my2 + ny2/m + n)
m = 7.28, n = 4.47
x1 = 7
x2 = 13
y1 = 5
y2 = 1
By point B = 7.28×1 + 4.47×7/7.28 + 4.47
(94.64 + 31.29/11.75, 7.28 + 22.35/11.75)
The three points AB and C are collinear
(125.93/11.75, 29.63/11.27)
(10.7, 2.6)
Example 5.
Find the values of a such that PQ = QR, where P, Q, and R are the points whose coordinates are (1, – 1), (1, 3), and (a, 8) respectively.
Solution:
We know that
Distance AB = √(x1 – x2)² + (y2 – y1)²
PQ = √(1−1)²+(−1−3)²
=√0+(−4)²
=√ 0+16
= √16
QR =√ (1−a)²+(3−8)²
= √(1−a)²+(−5)²
= √(1−a)²+25
Therefore, PQ = QR
√16 =√ (1−a)²+25
16 = (1 – a)² + 25
(1 – a)² = 16 – 25
(1 – a)² = -9
1 – a = ±3
a = 1 ±3
a = (4, -2)
FAQs on Distance and Section Formulae
1. Which is section formula?
Section formula is used to find the ratio in which a line segment is divided by a point internally or externally. It is used to find out the centroid, incenter, and excenters of a triangle.
2. What are the distance formula section formula and midpoint formula?
The distance formula is used to find the distance between two points in the plane. The Pythagorean Theorem, a2+b2=c2, is based on a right triangle where a and b are the lengths of the legs adjacent to the right angle, and c is the length of the hypotenuse.
3. Is the section formula and distance formula the same?
The distance formula is used to find the distance between two defined points on a graph. The section formula gives the coordinates of a point that divides the line joining two points in a ratio, internally or externally.