(1) The abscissa and ordinate of a given point are the distances of the point from y-axis and x-axis respectively.

(2) The coordinates of any point on x-axis are of the form (x,0).

(3) The coordinates of any point on y-axis are of the form (0,y).

(4) The distance between points P(x1,y1) and Q(x2,y2) is given by,
PQ=(x2−x1)2+(y2−y1)2−−−−−−−−−−−−−−−−−−√
For Example:
If P(−6,7) and Q(−1,−5), then distance between these two point is given as follow:
Here, x1=−6, y1=7
x2=−1,  y2=−5
Let O be the distance between two points (−6,7) and (−1,−5).
The distance between two point is given by
D=(x2−x1)2+(y2−y1)2−−−−−−−−−−−−−−−−−−√
=(−1−(−6))2+(−5−7)2−−−−−−−−−−−−−−−−−−−−−√
=(−1+6)2+(−12)2−−−−−−−−−−−−−−−−√
=(5)2+(−12)2−−−−−−−−−−−√
=25+144−−−−−−−√
=169−−−√
=13  units
Hence the distance between two points is 13 units.

(5) Distance of a point P(x,y) from the origin O(0,0) is given by OP=x2+y2−−−−−−√
For Example: P(−6,7) and O(0,0) is given then distance between them i.e. OP is calculated as follow:
OP=(x2−0)2+(y2−0)2−−−−−−−−−−−−−−−−−√
OP=x2+y2−−−−−−√
OP=62+72−−−−−−√
OP=85−−√

(6) The coordinates of the point which divides the join of points P(x1,y1) and Q(x2,y2) internally in the ration m:n are (mx2+nx1m+n,my2+ny1m+n)
For Example: Find the coordinates of the point which is divided the line segment joining (-1,3) and (4,-7) internally in the ratio 3:4.
Solution: Let the end points of AB be A(-1, 3) and B(4, -7).
(x1=−1, y1=3) and (x2=4, y2=−7)
Also, m=3 and n=4
Let P(x,y) be the required point, then by section formula, we have x=mx2+nx1m+n, y=my2+ny1m+n
⇒            x=3×4+4×(−1)3+4, y=3×(−7)+4×33+4
⇒            x=12−47, y=−21+127
⇒            x=87, y=−97
Hence, the required point is P(87,−97)

(7) The coordinates of the mid-point of the line segment joining the points P(x1,y1) and Q(x2,y2) are (x1+x22,y1+y22).
For Example: P(4,8) and Q(6,10)
Mid−Point=(4+62,8+102)
So mid-point of PQ is M(5,9).

(8) The coordinates of the centroid of triangle formed by the points A(x1,y1), B(x2,y2) and C(x3,y3) are (x1+x2+x33,y1+y2+y33)
For Example: Find centroid of triangle whose vertices are (1,4), (−1,−1), (3,−2)
Solution: We know that the coordinates of the centroid of a triangle whose angular points are (x1,y1), (x2,y2) and (x3,y3) are
(x1+x2+x33,y1+y2+y33)
So, the coordinates of the centroid of a triangle whose vertices are (1,4), (−1,−1) and (3,−2) are
(1−1+33,4−1−23)=(1,13)

(9) The are of triangle formed by the points A(x1,y1), B(x2,y2) and C(x3,y3) is 12|x1(y2−y3)+x2(y3−y1)+x3(y1−y2)| Or, 12|(x1y2+x2y3+x3y1)−(x1y3+x2y1+x3y2)|
For Example: Find area of triangle whose vertices are (6,3), (−3,5), (4,2)
Solution: Let A=(x1,y1)=(6,3), B=(x2,y2)=(−3,5) and C=(x3,y3)=(4,−2) be the given points
Area of ΔABC=12|{x1(y2−y3)+x2(y3−y1)+x3(y1−y2)}|
=12|{6(5−(−2))−3(−2−3)+4(3−5)}|
=12|{6×7+15−8}|
=12|57−8|
=492  sq.units

(10) If points A(x1,y1), B(x2,y2) and C(x3,y3) are collinear, then
x1(y2−y3)+x2(y3−y1)+x3(y1−y2)
For Example: Let A(2,5), B(4,6) and C(8,8) be the given points.
Three points are collinear if area enclosed by three points is zero.
Area of ΔABC=12|{x1(y2−y3)+x2(y3−y1)+x3(y1−y2)}|
=12|2(6−8)+4(8−5)+8(5−6)|
=12|2×(−2)+4×(3)+8×(−1)|
=12|−4+12−8|
=12|−12+12|
=0