Here let us have a look at all formula of coordinate geometry Class 10.
-
Distance Formula
-
Section Formula
-
Area of a Triangle
A detailed explanation of coordinate geometry class 10 all formulas are given below.
Coordinate Geometry Class 10 Formulas – Distance Formula
To calculate the distance between two points the distance formula is used. Making a triangle by using the Pythagorean theorem to find the length of the hypotenuse gives the distance formula. The distance between the two points in a triangle is called the hypotenuse.
The distance formula can also be used to calculate the lengths of all the sides of a polygon, the perimeter of polygons on a coordinate plane, the area of polygons, and several other things.
The distance formula is denoted by ‘d’.
Distance Formula Between 2 Points in a 2D Plane:
Consider 2 points P and Q having the 2D coordinates as (x1,y1) and (x2,y2) respectively.
Now the distance between these 2 points in the 2D plane is
[d = PQ = sqrt{(x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2}}]
Distance Formula Between 2 Points in a 2D Plane in Polar Coordinates:
Consider 2 points P and Q having the 2D polar coordinates as (r1,θ1) and (r2,θ2) respectively.
Now the distance between these 2 points in the 2D plane is
[d = PQ = sqrt{r_{1}^{2} + r_{2}^{2} – 2r_{1}r_{2} cos(theta_{1} – theta_{2})}]
Distance Formula Between 2 Points in a 3D Plane:
Consider 2 points P and Q having the 3D coordinates as (x1,y1,z1) and (x2,y2,z2) respectively.
Now the distance between these 2 points in the 3D plane is
[d = PQ = sqrt{(x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2} + (z_{2} – z_{1})^{2}}]
Distance Formula Between a Point and a Line
Consider a straight line Ax + By + C = 0 and point P having the 2D coorrdinates as (x1, y1).
Now the distance between the line Ax + By + C = 0 and a point P (x1, y1) is
[d = frac{|Ax + By + C|}{sqrt{A^{2} + B^{2}}}]
Distance Formula Between Two Parallel Points
Consider two parallel lines y = mx + c1 and y = mx + c2.
Now the distance between these two parallel lines is given by
[d = frac{|c_{1} – c_{2}|}{sqrt{A^{2} + B^{2}}}]
Problems on Distance formula
1. Find the distance between two points A and B which are having the 2D coordinates as (4, 8) and (3, 6) respectively.
Ans: The distance between these 2 points in the 2D plane is
[d = AB = sqrt{(x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2}}]
Here x1 = 4, x2 = 8, y1 = 3 and y2 = 6. Now substituting these values in the distance formula we get,
[d = AB = sqrt{(8 – 4)^{2} + (6 – 3)^{2}}]
[d= AB = sqrt{(4)^{2} + (3)^{2}}]
[d = AB = sqrt{16 + 9}]
[d = AB = sqrt{25}]
d = AB = 5 units
So the distance between 2 points A and B is 5 units.
Coordinate Geometry Formulas Class 10 – Section Formula
The Section formula is used in coordinate geometry to find the ratio in which a line segment is separated by a point, either internally or externally. It is used to determine the triangle’s centroid, incenter, and excenters.
Section Formula When the Line Segment Separated Internally by a Point
Consider a point P (x, y) dividing the line segment joining the points A(x1,y1) and B(x2,y2) internally in the ratio m:n.
Now the section formula for a line segment separated internally by a point is:
[P(x, y) = (frac{mx_{2} + nx_{1}}{m + n}, frac{ny_{2} + ny_{1}}{m + n})]
Section Formula When the Line Segment Separated Externally by a Point
Consider a point P (x, y) dividing the line segment joining the points A(x1, y1) and B(x2, y2) externally in the ratio m:n.
Now the section formula for a line segment separated externally by a point is:
[P(x, y) = (frac{mx_{2} – nx_{1}}{m – n}, frac{ny_{2} – ny_{1}}{m – n})]
Section Formula When Line Segment Separated Internally by a Point in the Ratio K:1
Consider a point P (x, y) dividing the line segment joining the points A(x1,y1) and B(x2,y2) internally in the ratio K:1.
Now the section formula for a line segment separated internally by a point is:
[P(x, y) = (frac{Kx_{2} + x_{1}}{K + 1}, frac{Ky_{2} + y_{1}}{K + 1})]
Section Formula When Line Segment Separated Externally by a Point in the Ratio K:1
Consider a point P (x, y) dividing the line segment joining the points A(x1,y1) and B(x2,y2) externally in the ratio K:1.
Now the section formula for a line segment separated externally by a point is:
[P(x, y) = (frac{Kx_{2} – x_{1}}{K – 1}, frac{Ky_{2} – y_{1}}{K – 1})]
Problems on Section Formula
1. Two points A (5,8) and B (3,6) on the line segment are separating the point P (x,y) internally in the ratio 2:4. By using the section formula find the ratio in which a line segment is separated by a point.
Ans: The section formula for a line segment separated internally by a point is given by the formula:
[P(x, y) = (frac{mx_{2} + nx_{1}}{m + n}, frac{ny_{2} + ny_{1}}{m + n})]
Here x1 = 5, x2 = 8, y1 = 3 and y2 = 6, m=2, n=4. Now substituting these values in the section formula we get,
[P(x, y) = (frac{(2 times 8) + (4 times 5)}{2 + 4}, frac{(2 times 6) + (4 times 3)}{2 + 4})]
[P(x, y) = (frac{16 + 20}{6}, frac{12 + 12}{6}]
[P(x, y) = (frac{36}{6}, frac{24}{6})]
Px, y = 6, 4
The ratio in which the line segment separated by a point internally is (6,4).
2. Two points A (5,8) and B (3,5) on the line segment are separating the point P (x,y) externally in the ratio 2:4. By using the section formula find the ratio in which a line segment is separated by a point.
Ans: the section formula for a line segment separated externally by a point is
[P(x, y) = (frac{mx_{2} – nx_{1}}{m – n}, frac{ny_{2} – ny_{1}}{m – n})]
Here x1 = 5, x2 = 8, y1 = 3 and y2 = 5, m=2, n=4. Now substituting these values in the section formula we get,
[P(x, y) = (frac{(2 times 8) – (4 times 5)}{2 – 4}, frac{(2 times 5) – (4 times 3)}{2 – 4})]
[P(x, y) = (frac{16 – 20}{-2}, frac{10 – 12}{-2}]
[P(x, y) = (frac{-4}{-2}, frac{-2}{-2})]
P(x, y) = (2, 1)
The ratio in which the line segment separated by a point externally is (2, 1).
Formulas of Coordinate Geometry Class 10 – Area of a Triangle
The area of a triangle is defined as the total area enclosed by the triangle’s three sides.
Here in this coordinate geometry all formula Class 10 we will discuss the area of the triangle of three different triangles namely right-angled triangle, equilateral triangle, and isosceles triangle.
Area of a Right Angled Triangle
A right-angled triangle is one in which one of the angles is the right angle. The hypotenuse is the hand opposite the right angle. Legs are the sides that are adjacent to the right angle.
The area of the right-angled triangle is:
A = ½ × Base × Height
Area of an Equilateral Triangle
An equilateral triangle is one in which all three sides are the same length and all three internal angles are congruent and 60°each.
The area of an equilateral triangle is:
[A = frac{sqrt{3} times a^{2}}{4}]
Area of an Isosceles Triangle
A triangle with two sides of equal length is called an isosceles triangle.
The area of an isosceles triangle is:
A = ½ × Base × Height
Area of Triangle With all 3 Sides Different
When all 3 sides of a triangle are different, then to find the area of the triangle we will use Heron’s formula.
When the lengths of all three sides are known, Heron’s formula, named after Hero of Alexandria, measures the area of a triangle. Unlike other triangle area formulas, no angles or other distances in the triangle must be determined first.
Heron’s formula to calculate the area of the triangle is:
[A = sqrt{s(s – a)(s – b)(s – c)}]
Where ‘s’ is the semi perimeter of the triangle and a, b and c are the sides of the triangle.
Semi perimeter of the triangle is given by:
S = (a + b + c)/2
Problems on Area of Triangle
1. If the base of the right-angled triangle is 4 cm and height is 6 cm. Find the area of the triangle.
Ans: The area of the right-angled triangle is:
A = ½ × Base × Height
Here Base = 4 cm and Height = 6 cm. Substituting these values in the formula we get,
A = ½ × 4 × 6
A = ½ × 24
A = 12 cm2.
Therefore the area of the right-angled triangle is 12 cm2.
2. If the three sides of a triangle are having the same length equal to 6 cm. Find the area of the triangle and name the type of triangle.
Ans: If all three sides are the same length then the triangle is an equilateral triangle.
The area of an equilateral triangle is:
[A = frac{sqrt{3} times a^{2}}{4}]
Here ‘a’ is the sides of the triangle given as 6cm. So substituting this value we get
[A = frac{sqrt{3} times 6^{2}}{4}]
[A = frac{sqrt{3} times 36}{4}]
[A = sqrt{3} times 9]
[A = 9sqrt{3}] cm2
Therefore the area of the equilateral triangle is [9sqrt{3}] cm2.
Conclusion
One of the most important and exciting concepts in mathematics is coordinate geometry. By the use of graphs of lines and curves, it connects algebra and geometry. This allows algebraic solutions to geometric problems and offers geometric insights into algebra. These coordinate geometry Class 10 formulas will help students to understand the basic concepts of geometrical structures and to apply them in our daily uses too.