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Algebraic Geometry

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Algebraic geometry is the combination of linear algebra and algebra. Algebraic geometry Combines these two fields of mathematics by studying systems of polynomial equations in several variables. Algebraic geometry occupies a central place in modern mathematics and has multiple conceptual connections with such diverse fields as complex analysis, topology and number theory.

It is important to study the intrinsic properties of a system of equations that leads into the deepest areas in mathematics. The fundamental objects of study in algebraic geometry are algebraic varieties, in algebraic varieties we study plane algebraic curves, including lines, circles, parabolas, quartic curves etc., Here the point of interest is like singular points, inflection points and points at infinity. 

How to Solve Algebra Problems?

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Below we describe some algebraic equations with brief description.

Quadratic equation

A quadratic equation always represents a parabola. Every quadratic form of equation when graphically represented, forms a parabola. When a ball is thrown or kicked, it follows a parabola.
The equation of parabola is can be written as
y = ax$^{2}$ + bx + c where a $\neq$ 0
Where,
a, b and c are constants.
(x, y) is an arbitrary point lying on the parabola.
a ≠ 0. If a equals zero, then equation will be no more a quadratic equation. Hence it will not represent a parabola.

Distance formula
In analytical geometry, distance formula measures the distance between two points whose endpoints coordinates are specified as ($x_{1}, y_{1}$) and ($x_{2}, y_{2}$)
The distance formula is given below:
d = $\sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2})}$
d is the distance between two points.

Distance From a Point to a Line Using Line Equation

The distance between a line and a point which is not on the line can be found by using a application of lines. This distance is defined as the length of the perpendicular line segment joining the point and the line.

The Distance between the point (x1, y1) and the line Ax + By + C = 0 is given as

d = $\frac{|Ax_{1} + By_{1} + C|}{\sqrt{A^{2} + B^{2}}}$

Centroid of a triangle
Centroid of a triangle is the point where the three medians of the triangle meet. Coordinates of centroid are ($x_{1}, y_{1}$), ($x_{2}, y_{2}$) and ($x_{3}, y_{3}$), formula for the centroid of a triangle is given as

x = $\frac{x_{1}+x_{2}+x_{3}}{3}$  and  y = $\frac{y_{1}+y_{2}+y_{3}}{3}$

Formula to find slope point

Slope intercept formula for a line

 y = mx + b
where m : slope of the line
b : y-intercept

Formula for slope point of a line

y - $y_{1}$ = m (x - $x_{1}$)
Point on the line : ($x_{1}$, $y_{1}$)
m : slope of the line.

Algebraic Geometry Solutions

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Solved Examples

Question 1: Consider a parabola: y$^2$ - 16x = 0. Determine its vertex and focus.
Solution:
 
y$^{2}$ = 16x
Comparing it with standard parabola equation, y$^2$ = 4ax:
Vertex = (0, 0)
4a = 16
a = 4
Focus = (a, 0) = (4, 0)
 

Question 2: Find the distance between the points (1, 5) and (3, -2)
Solution:
 
In the given problem x$_{1}$ = 1, $y_{1}$ = 5 and x$_{2}$ = 3, y$_{2}$ = -2
The distance formula is
d = $\sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2})}$
= $\sqrt{(3-1)^{2}+(-2-5)^{2}}$
= $\sqrt{4+49}$
= $\sqrt{53}$
= 7.28
 

Question 3: The Distance between the point (5, 2) and the line y = 0.25x + 5
Solution:
 
Converting the line in the general form we get,
-0.25 x + y - 5 = 0

Using the formula we get

d = $\frac{|-0.25(5) + 1 (2) + (-5)|}{\sqrt(-0.25)^{2} + 1^{2}}$

= $\frac{| -1.25 + 2 - 5|}{1.033}$

= 4.114

Therefore, the length of the line segment is the distance what we are looking for and the distance from point to the line is 4.114
 

Question 4: Solve the given equation of the tangent line for the curve f(x) = x$^4$ + 10 at x = 3.
Solution:
 
Given f(x) = x$^4$ + 10
f'(x) = 4x$^3$

At x = 3, slope of the tangent line is f'(3)
f'(3) = 4(3)$^3$
f'(3) = 108 (slope)

As $x_{1}$ is known, find $y_{1}$ by substituting x = 3 in f(x)

f(3) = 3$^4$ + 10
f(3) = 91

Now we'll be able to find equation of a tangent line with the help of slope point formula.
At  the point (3, 91), f'(3) = 108 equation of tangent line is
y - y$_1$ = m (x - $x_1$)
y - 91 = 108 (x - 3)
y - 91 = 108x - 324
y = 108 x - 233