The coordinate geometry is also called a coordinate plane. It enables a plane in perpendicular axes : x and y in two-dimensional form. The horizontal direction indicates the x-axis while the vertical direction y-axis indicates of the plane. In these points are indicated by their positions along the x and y-axes.
Coordinate Proof and Quadrilaterals
Proof: - Positioning a Square
Position and label a square with sides n units long on the coordinate plane.
Let A, B, C, and D be vertices of the square.
Place A at the origin. As a result, AB is on the
positive x-axis and AD is on the y-axis. Label
the vertices A, B, C, and D.
The y-coordinate of B is 0 because the vertex is
on the x-axis. Since the side length is n,
the x-coordinate is n.
D is on the y-axis so the x-coordinate is 0.
The y-coordinate is 0 + n or n.
The x-coordinate of C is also n. The y-coordinate is
0 + n or n because the side BC is n units long.
Sample Problems of Geometry Quadrilaterals:
Q 1: Name the type of the quadrilaterals formed in coordinate geometry, if any,by a following points,and give reasons for your answer:
(i) (-1,-2)(1,0),(-1,2),(-3,0)
Solution for coordinate geometry using quadrilaterals: Let A (-1,-2),B(-1,2)and D(-3,0) be the four given points
Then using distance formula,we have
AB=v(1+1)2+(0+2)2 =v4+4=v8= 2v2
BC=v(-1-1)2+(2-0)= v4+4=v8=2v2
CD=v(-3+1)2+(0-2)2 =v4+4=v8=2v2
DA=v(-1+3)2+(-2+0)2 =v4+4=v8=2v2
AC= v(-1+1)2+(2+2)2 =v0+16=4
and BD= v(-3-1)2+(0-0)2 =v16=4
Hence Four sides of coordinate geometry quadrilaterals are equaland diagonal ACand BD are also equal coordinate geometry Quadrilaterals ABCD is a square
(ii) (4,5),(7,6),(4,3),(1,2)
sol : Let A(4,5),B(7,6),C(4,3)and D(1,2) be the given points.then,
AB=v(7-4)2+(6-5)2 =v9+1=v10
BC=v(4-7)2+(3-6)2 = v9+9 = v18=3v2
CD=v(1-4)2+(2-3)2 =v9+1=v10
DA=v(4-1)2+(5-2)2 =v9+9=v18=3v2
AC=v(4-4)2+(3-5)2 = v0+4=2
and BD=v(1-7)2+(2-6)2 =v36+16=v52=2v13
clearly AB=CD,BC=ADand AC?BD
ABCD is a paralelogram
Q 2 : Is(1,2),(4,y),(x,6)and(3,5) are vertices of the parallelogram taken in order,find xand y.
sol : LetA(1,2),B(4,y),C(x,6)andD(3,5) be the vertices of a parallelogramABCD
since this diagonals of the parallelogram bisect each other
(x+1/2,6+2/2)=(3+4/2,5+y/2)
=> x+1/2=7/2 =>x+1=7 x=6
and 4=5+y/2 =>5+y y=8-5=3
Hence x=6and y=3
Q 3 : Find the area of the triangle whose vertices are
(-5,-1),(3,-5),(5,2)
Sol : let A(x1,y1)=(-5,-1)
B(x2,y2)=(3,-5)
c(x3,y3)=(5,2)
area of ?ABC=1/2[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]
= 1/2[-5(-5-2)+3(2+1)+5(-1+5)]
= 1/2(35+9+20)=1/2 x64=32sq.units
practice problem of Geometry quadrilaterals:
1.Find the area of a rhombus if its vertices are (2,0),(6,7),(-2,5)and(-4,-5)taken in order
2.find the point on the x-axis which is equidistant form(3,-6)and(-1,8)
