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. |

**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 $\neq$ 0Where,

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 (x

_{1}, y

_{1}) 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.

### 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