Chapter 7: Coordinate Geometry

Let the vertices be (x,y)

Distance between (x,y) & (4,3) is = √((x-4)² + (y-3)²)……(1)

Distance between (x,y) & (-4,3) is = √((x+4)² + (y-3)²)……(2)

Distance between (4,3) &(-4,3) is =√((4+4)² + (3-3)²) = √(8)²=8

According to the question,

Equation (1)=(2)

(x-4)²=(x+4)²

x²-8x+16=x²+8x+16

16x=0

x=0

Also, equation (1)=8

(x-4)²+(y-3)²=64……… (3)

Substituting the value of x in (3)

Then (0-4)²+(y-3)²=64

(y-3)²=64-16

(y-3)²=48

y-3=(+)4√3

y=3(+) 4√3

Neglect y = 3+4√3 as if y = 3+4√3 then origin cannot interior of triangle

Therefore, the third vertex = (0, 3-4√3)

According to the question,

The three vertices of a parallelogram ABCD are A (6, 1), B (8, 2) and C (9, 4)

Let the fourth vertex of parallelogram = (x, y),

We know that, diagonals of a parallelogram bisect each other

Since, mid – point of a line segment joining the points (x₁, y₁) and (x₂, y₂) is given by,

∵ Area of ΔABC with vertices (x₁, y₁), (x₂, y₂) and (x₃, y₃);

= ½[x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]

∴ Area of ΔADE with vertices A (6, 1), D (7, 3) and E (8, (7/2))

= – ¾ but area can’t be negative

Hence, the required area of ΔADE is ¾ sq. units

According to the question,

The vertices of ΔABC = A, B and C

Coordinates of A, B and C = A(x₁, y₁), B(x₂, y₂), C(x₃, y₃)

(i) As per given information D is the mid – point of BC and it bisect the line into two equal parts.

Coordinates of the mid – point of BC;

(ii) Let the coordinates of a point P be (x, y)

Given,

The ratio in which the point P(x, y), divide the line joining,

A(x₁, y₁) and D((x₂+x₃)/2 , (y₂+y₃)/2)) = 2:1

Then,

Coordinates of P =

(iii) Let the coordinates of a point Q be (p, q)

Given,

The point Q (p, q),

Now,

Let the coordinates of a point E be (⍺, β)

Given,

(iv) Coordinate of the centroid of the ΔABC;

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