# Geometry Formulas for Class 11

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## 3D Geometry Formulas Class 11

In this chapter, the Three-Dimensional Geometry Formulas Class 11 Notes are gathered in a systematic manner to help students get rid of confusion regarding the course content and concepts since CBSE keeps on updating the course every year. The formulae list is inclusive of all coordinate geometry class 11 formulas which make it super simple for students to study, revise and memorize the chapter. So, let’s get started with introduction to 3d Geometry Class 11 notes to kick start you with an effective preparation for your examinations.

### Introduction to 3d Geometry Formulas Class 11

In order to detect the position of a geometric object in a line, you just need the distance of the object from the point of reference. However, for the purpose of locating the position of a point in a plane, you need two bisecting conjointly perpendicular lines in the plane. These lines in the plane are referred to as the coordinate axes and the two numbers are referred as the coordinates of the point in regards to the axes. See the figure below and locate point O that’s what we call the origin of the coordinate system. The three coordinate planes essentially divide the space into eight parts called the octants.

### Coordinate Geometry Formulas Class 11

A. Rectangular Cartesian Coordinates:

(i) If the starting line and pole of the polar system coincides individually with the point of origin and positive x-axis of the Cartesian plane and (x, y), (r, θ) be the polar coordinates and cartesian respectively of a point P on the plane then,

x = r cos θ

y = r sin θ and r = $\sqrt{(x_{2} + y_{2})}$,

θ = $tan^{-1}(y/x)$.

(ii) This states the distance between two allotted points M (x1, y1) and N (x2, y2) is given by $MN = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$.

(iii) The coordinates of the centroid of the triangle created by connecting the points (x1, y1) , (x2, y2) and (x3, y3) are $\frac{(x_{1} + x_{2} + x_{3})}{3} , \frac{(y_{1} + y_{2} + y_{3})}{3}$.

(iv) The area of a triangle created by connecting the points (x1, y1), (x2, y2) and (x3, y3) is given by ½ $|y_{1} (x_{2} - x_{3}) + y_{2} (x_{3} - x_{1}) + y_{3} (x_{1} - x_{2})|$ sq. Units.

### B. Circle Geometry Formulas

The equation of the circle that has midpoint at the origin and radius a units is x2 + y2 = m2

The circle’s parametric equation is x = a cos θ and y = a sin θ, θ being the parameter.

The following table gives some important circles and its variations formulas.

C. Coordinate Geometry Formulas

The Cartesian plane or coordinate plane is a fundamental theory for coordinate geometry. It delineates a 2d plane with respect to two perpendicular axes: x and y. The x-axis describes the horizontal direction while the y-axis describes the vertical direction of the coordinate system. In the coordinate plane system, points are indicated by their locations along the x and y-axes.

The formulas given below in a tabular form provides some important coordinate geometry formulas and how to use them for solutions.

 Slope Formula Slope, $m = \frac{\text{change in y}}{\text{change in x}} = \frac{rise}{run} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$Parallel lines have equal slopes.The slopes of perpendicular lines are opposite reciprocals of each other. General Formula $Ax + By = C$ Slope Intercept Formula $y = mx + b$Where, m is the slope and b is the y-intercept Point Slope Formula $(y - y_{1}) = m(x - x_{1})$Where, m is the slope Midpoint Formula $(\frac{x_{2} + x_{1}}{2}, \frac{y_{1} + y_{2}}{2})$ Distance Formula $\sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$

D. Parabola

(i) Elementary equation of parabola is given as; $y^{2} = 4ax$. Its axis is x-axis and its vertex is the origin.

(ii) Other kinds of the equations of parabola are as given:

(a) $x^{2} = 4ay$ (axis is y-axis and vertex is the origin)

(b) $(y - \beta)^{2} = 4a (x - \alpha)$ [axis is parallel to x-axis and the vertex is at (α, β)]

(c) $(x - \alpha)^{2} = 4a(y - \beta)$ [axis is parallel to y-axis and vertex is at ( a, β)]

(iii) The parabola’s parametric equations are given by; $y^{2} = 4ax$ are $x = at^{2}$ , y = 2at, t being the parameter.

### Solved Examples

Example1:

Prove the type of triangle of the complex numbers m1, m2, and m3 respectively the points P, Q, R on the Argand plane and = 0, then what type of triangle is ΔPQR.

Solution1:

Given that, we have

$\begin{bmatrix}m_{1} &m_{2} &1 \\m_{2} &m_{3} &1 \\m_{3} &m_{1} &1 \end{bmatrix}$

= $m_{1} (m_{3} - m_{1}) - m_{2} (m_{2} - m_{3}) + 1 (m_{2}m_{1} - m_{3}^{2}) = 0$

= $m_{1}m_{2} + m_{2}m_{3} + m_{1}m_{3} - m_{1}^{2} - m_{2}^{2} - m_{3}^{2} = 0$

This occurs to be the condition for vertices of an equilateral triangle.

Thus ΔPQR is an equilateral triangle.

Example 2:

A root of $y^{5} - 32 = 0$ is located in quadrant II. Express this root in polar form.

Solution2:

We already have:  $x^{5} - 32 = 0$

This states: $y^{5} = 2^{5} e^{i2\pi n}$

So we obtain $y = \frac{2 e^{2\pi n}}{5}$

Given that the root is in the second quadrant, thus we have our answer y = 2(cos (144) + i sin (144)).

Q1: Are Class 11 Math Formulas and Concepts Sufficient to Score Good Marks in the National Aptitude Test in Architecture (NATA)?

Ans: Though there may be people that might tell you the maths section of NATA is not that difficult to score, underrating it will be completely wrong. Besides, knowing Class 11 and 12 maths formulas, you must know all the concepts, laws and properties. That said, to get through the NATA examination you should practice plenty of maths mock questions.

Q2: What is Covered in the Conic Sections of Maths Formulas for Class 11?

Ans: In Euclidean geometry, a circle is a spherical figure where all the points in a plane are positioned at an equal distance from the fixed point on a given plane. Following are the most important formulas covered in conic section of math class 11:

• The equation of the circle with the median point (h, k) and radius (r) is written as (x – h)2 + (y – k)2 = r2

• The equation of the parabola that has foci at point (a, 0) where a > 0 and directrix ‘x’ = – a is written as: y2 = 4ax

• The equation of a hyperbola with focal point on the x-axis is given by x2/a2 − y2/b2 = 1

• The equation of an ellipse with focal point on the x-axis is x2/a2 + y2/b2 = 1

• Length of the latus rectum of the hyperbola [x2/a2 − y2/b2 = 1] is written as: 2b2/a

• Length of the latus rectum of the ellipse [x2/a2 + y2/b2 = 1] is written as: 2b2/a