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2-D coordinate system- Analytic geometry Formulas

2-D coordinate system

A two-dimensional Cartesian coordinate system is formed by two mutually perpendicular axes. The axes intersect at the point O, which is called the origin. In the right-handed system, one of the axes (x-axis) is directed to the right, the other y-axis is directed vertically upwards.

The coordinates of any point on the xy-plane are determined by two real numbers x and y, which are orthogonal projections of the points on the respective axes. The x-coordinate of the point is called the abscissa of the point, and the y-coordinate is called its ordinate.

Distance between two points:

Consider two points A(x1,y1) and B(x2,y2) in an XY plane.

Then distance formula is given by,

AB=(x2–x1)2+(y2–y1)2 AB= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Example:

Find the distance between the points A(2,-3) and B(5,1).

AB=(5–2)2+(1–(–3))2 AB= \sqrt{(5-2)^2+(1-(-3))^2}
AB=32+42=25 AB= \sqrt{3^2+4^2} = \sqrt{25}
AB=5 units AB= 5\ \text{units}

Section Formula:

Let A(x1 , y1) and B(x2, y2) be the two pints on the XY plane. And P( x, y) be any point on segment AB such that AP: PB = m:n.

Condition 1: If the point P divides segment AB internally in the ratio m:n.

Then the (x,y) coordinates of P are given by:

mx2+nx1m+n,my2+ny1m+n \left[{mx_2+nx_1 \over m+n},{my_2+ny_1 \over m+n}\right]

Condition 2: If the point P divided the line segment AB externally in the ratio m:n.

Then the ( x, y) coordinates of P are given by

mx2–nx1m–n,my2–ny1m–n \left[{mx_2-nx_1 \over m-n},{my_2-ny_1 \over m-n}\right]

Condition 3: If the point P divided the line segment AB in the equal ratio i.e m = n or m:n = 1:1 then the ( x, y) coordinates of P are given by

P(x,y)=x1+x22,y1+y22 P(x,y)= \left( {x_1+x_2 \over 2},{y_1+y_2 \over 2} \right)

Example:

P (4, 5) and Q(7, – 1) are two given points and the point R divides the line-segment PQ externally in the ratio 4:3. Find the coordinates of Y.

Given that, P(4,5)=(x1,y1), Q(7,-1)=(x2,y2)

Point R divides the segment PQ in the ratio 4:3,

hence m=4, n=3

Since given that the point R divides the segment externally we use the section formula for external division,

$$ R= \left\lbrace \left[ {mx_2-nx_1 \over m-n} \right] , \left[ {my_2-ny_1 \over m-n} \right] \right\rbrace $$
$$ R= \left\lbrace \left[ {(4 x 7) – (3 x 4) \over (4 – 3) } \right] , \left[ {(4 x -1) – (3 x 5) \over(4 – 3)} \right] \right\rbrace $$
$$ R= \left\lbrace \left[ {(28 – 12) \over 1} \right] , \left[ {(-4 – 15) \over 1} \right] \right\rbrace $$
R=\left{16,–19\right} R= \left\lbrace 16,-19 \right\rbrace

The coordinates for the point R which divides the segment PQ externally are (16,-19)

Area of a Triangle:

The area of a triangle whose coordinates of vertices are (x1, y1) , (x2, y2) and (x3, y3)

The area of a triangle ABC whose vertices are A(x1, y1) , B(x2, y2) and C(x3, y3) is given by

Area of Triangle=12[x1(y2–y3)+x2(y3–y1)+x3(y1–y2)] Area \ of \ Triangle = \frac 12 [x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]

Note:

  • To find the area of a polygon we divide it in triangles and take the numerical value of the area of each of the triangles.
  • Three points A(x1, y1) , B(x2, y2) and C(x3, y3) are collinear if and only if Area of triangle = 0
12[x1(y2–y3)+x2(y3–y1)+x3(y1–y2)]=0 \frac 12 [x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)] = 0

Example:

Find the area of the LMN whose vertices are L(3, 2), M(4, 2) and N(3, 5)

A=12[3(2–5)+4(5–2)+3(2–2)] A = \frac 12 [3(2 – 5) + 4(5 – 2) + 3(2 – 2)]
A=12[3×(–3)+4×3] A = \frac 12 [3 \times (-3) + 4 \times 3]
A=12[–9+12] A = \frac 12 [- 9 + 12]
A=12×3 A = \frac 12 \times 3
A=32 square units. A = \frac 32 \ square \ units.

Slope of a Straight Line (m):

The slope of a line denotes the steepness of a line. Slope is the tangent of the angle made by straight line with the positive direction of X-axis.

m=(y2–y1)(x2–x1)=tanθ m= \sqrt {(y_2-y_1) \over (x_2-x_1)} = tan \theta

Example:

Find the slope of the line passing through the coordinates A(1, 9) and B(3, -5)?

slope m=tanθ=(y2–y1)(x2–x1) slope \ m = tan \theta = \sqrt {(y_2-y_1) \over (x_2-x_1)}
m=–5–93–1=–142=–7 m= \frac {-5-9}{3-1} = – \frac {14}{2} = -7

Thus, slope of the given line will be -7.

Equation of a Straight Line:

The equation of line parallel to X-axis is given as y = d

The equation of line parallel to Y-axis is given as x = c

There are many forms a straight line can be represented:

    • Slope form:

The equation of straight-line having slope m is given as y = mx + c, where c is the intercept of a straight line on Y-axis.

    • Intercept form:

The intercept form of a straight line having a and b as the intercepts on X-axis and Y-axis respectively is given as

xa+yb=1 \frac xa + \frac yb = 1
    • Point slope form:

The equation of line passing through point (x1, y1) and having slope m is given as

y–y1=m(x–x1) y-y1=m(x-x_1)
    • Two-point form:

The equation of line passing through points (x1, y1) and (x2, y2) is given as

y–y1=(y2–y1)(x2–x1)(x–x1) y-y1={(y_2-y_1) \over (x_2-x_1) }(x-x_1)
    • General form:

ax + by + c = 0 where a, b, c are constants and a, b are not both zero.

    • Normal Form:

Normal form of any straight line is defined by the length of the perpendicular (p) from the origin to the line and angle (θ) which perpendicular makes with the positive direction of X-axis. Any straight line can be represented in normal form as

xcosθ+ysinθ=p xcos \theta + y sin \theta = p

Angles between two lines

Let “θ” be the angle between these two lines, we can then represent the angle between them as

tanθ=(m1–m2)1+m1m2 tan \theta = \left[ {(m_1-m_2) \over 1+m_1m_2} \right]

Midpoint Theorem:

Consider the same points A and B, having coordinates to be (x1,y1) and (x2,y2) respectively. Let M(x,y) be the midpoint of lying on the line connecting these two points A and B. The coordinates of this point are

M(x,y)=12(x1+x2) ,12(y1+y2) M(x,y)= \left[ \frac 12 (x_1+x_2) \ , \frac 12 (y_1+y_2) \right]

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