**Origin:**Through a point O, referred to as the**origin,**we take two mutually perpendicular lines XOX’ and YOY’ and call them x and y axes respectively.**Abscissa & Ordinate:**The position of a point is completely determined with reference to these axes by means of an ordered pair of real numbers (x, y) called the**coordinates**of P where | x | and | y | are the distances of the point P from the y-axis and the x-axis respectively, x is called the x-coordinate or the**abscissa**of P and y is called the y-coordinate or the**ordinate**of the point P.

The distance of the point P (x, y) from the origin O (0, 0) is given by,
**OP = √((x – 0) ^{2}+ (y – 0)^{2}), i.e. OP = √(x² + y²)**

Then, (x

PQ = √((x

= √((2 – (-4))

= √(6

= √(36 + 144)

= √180

= 6√5

x = ((-5) + 7)/2 =1 and y = ((4 + (-8)) = -2

Hence, the required point is M (1, -2).

given m:n = 2:3 and (x

now for M(x,y)

x = (mx

= (2×7 + 3×5)/(2 + 3)

= 29/5

y = (my

= (2×8 + 3×4)/(2 + 3)

= 40/5

= 8

So, the required point M (x,y) is (29/5, 8)

Slope (m) = (y

= (8-1)/(7-5)

= 7/2

Then,

(x

Area of △ABC = |1/2 {x

= 1/2 |2(-1 – 6) + 3(6 – 7) – 5(7 +1)|

= 1/2| -14 – 3 – 40|

= 1/2|-57|

= 57/2

= 28.5 sq units.

Then, (x

∴ x

= (-1) (7 – 10) + 5(10 -1) + 8(1 – 7)

= (3 + 45 – 48)

= 0

Hence, the given points are collinear.

Let A (2, 7), B (3, -1) and C (-5, 6) be the vertices of the given △ABC.

(x

Centroid (x,y) = {(x

= {2+3+(-5)}/3 , {7+(-1)+6}/3

= (0 , 12)

Then, (x

∴ a (c + a – a – b) + b (a + b – b – c) + c (b + c – c – a)

= a (c – b) + b (a – c)+ c (b – a)

= 0

Hence, the given points are collinear.

Area of △ABC = 1/2|{(-4).(-5 + 2) -3(-2 + 2) + 3(-2 + 5)}|

= 21/2 sq units.

Area of △ACD = 1/2|{ (-4).(-2 – 3) + 3(3 + 2) + 2(-2 + 2}|

= 35/2

Area of quad. ABCD = 21/2 + 35/2 sq units = 28 sq units.

Then, P divides AB in the ratio 1 : 2

So, the coordinates of P are

P ((1 x 4 + 2 x (-5)/1 + 2, ((1 x (-3) + 2 x 6)/1 + 2) i.e., P (-2, 3).

Also, Q divides AB in the ratio 2:1.

So, the coordinates of Q are

Q ((2 x 4 + 1 x (-5)/2 + 1, (2x(-3) +1 x 6/1 + 2)) i.e., Q (1, 0)

Q (1, 0).

Then, OP = √((6 – 0)

= √(6

= √72

= 6√2

- GPS and Map Navigation uses coordinate geometry to mark our location.
- It used by airline industry to plot the trajectory of a plane.
- Latitude and Longitude system uses coordinate geometry to work.
- It is also used for military purposes.

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