Coordinate Geometry

Solution NCERT Exercise 7.2 part2

Question (6) If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.

Solution

Given, vertices of parallelogram are (1, 2), (4, y), (x, 6) and (3, 5)

Thus, x and y = ?

Let the given vertices are A(1, 2), B(4, y), C(x, 6) and D(3, 5)

Since, the given vertices are of a parallelogram, thus diagonals each other at mid point.

This means diagonals cut each other in the ratio of 1:1

We know that, If the ratio in which a point P divides the line AB is k:1, then the coordinates of the point P will be `((kx_2+x_1)/(k+1), (ky_2+y_1)/(k+1))`

Calculation of coordinate of mid point O(a, b) for line segement AC

Here, we have,

x₁ = 1, x₂ = x

And, y₁ = 2, and y₂ = 6

And, ratio k: 1 = 1:1

Thus, `a=((1xx\x)+1)/(1+1)`

`a=(x+1)/2` - - - - (i)

And, `b=((1xx6)+2)/(1+1)`

`=>b=(6+2)/2=8/2`

`=>b = 4` - - - - - (ii)

Calculation of coordinate of mid-point O(a, b) for line segment BD

Here we have x₁ = 4, x₂ = 3

And, y₁ = y, and y₂ = 5

And, ratio k: 1 = 1:1

Thus, `a=(1xx3+4)/(1+1)`

`=>a=(3+4)/2`

`=>a=7/2` - - - - - (iii)

And, `b=((1xx5)+y)/(1+1)`

`=>b = (5+y)/2` - - - - - (iv)

Since, mid point is common, thus from equation (i) and (iii)

`(x+1)/2=7/2`

After cross multiplication, we ge

2(x + 1) = 14

⇒ 2x + 2 = 14

⇒ 2x = 14 – 2

⇒ 2x = 12

Thus, `x=12/2`

⇒ x = 6

And, from equation (ii) and (iv)

`4=(5+y)/2`

After cross multiplication, we get

4 × 2 = 5 + y

⇒ 8 = 5 + y

⇒ 8 – 5 = y

⇒ y = 3

Thus, x = 6 and y = 3 Answer

Question (7) Find the coordinate of a point A, where AB is the diameter of a circle whose centre is (2, –3) and B(1, 4).

Solution:

Given, coordinates of centre and one end point of a diameter are (2, –3) and B(1, 4)

Thus, coordinate of another end point A of diameter =?

Let, coordinate of A is x, y

Since, coordinate of mid point is (2, –3)

Thus, `2 =(x+1)/2`

After cross multiplication, we get

2 × 2 = x + 1

⇒ 4 = x + 1

Thus, x = 4 – 1 = 3

Thus, x = 3

And, `-3=(y+4)/2`

`=>-3=(y+4)/2`

After cross multiplication, we get

–3 × 2 = y + 4

⇒ –6 = y + 4

⇒ y = –6 – 4 = –10

Thus, coordinate of A(3, –10) Answer

Question (8) If A and B are (–2, –2) and (2, –4), respectively, find the coordinate of P such that AP = `3/7`AB and P lies on the line segment AB.

Solution

Given, A(–2, 2) and B(2, –4)

And, AP =`3/7` AB and P lies on the line segment AB

Then coordinate of P =?

Since, AP = `3/7` AB

Thus, `(AP)/(AB) = 3/7`

Clearly, AP = 3

Thus, PB = 4

`=>(AP)/(PB) = 3/4`

Thus, P divides the line segment AB in the ratio of 3:4

We know that,

According to section formula, if the coordinates of the point P(x, y) divides the line segment joining the points A(x₁, y₁) and B(x₂, y₂), internally, in the ratio m₁:m₂ , then

`x=(m_1x_2+m_2x_1)/(m_1+m_2)` and `y=(m_1y_2+m_2y_1)/(m_1+m_2)`

Here we have, x₁=–2, x₂ = 2

And, y₁ = –2, y₂ = –4

And, m₁ = 3, m₂ = 4

Let, P(x, y)

Thus, by using section formula of coordinate geometry,

`x=((3xx2)+(4xx(-2)))/(3+4)`

`=(6+(-8))/7`

`=(6-8)/7`

`=>x=(-2)/7`

And, `y=((3xx(-4))+(4xx(-2)))/(3+4)`

`=(-12+(-8))/7`

` = (-12-8)/7`

`=>y = (-20)/7`

Thus, coordinates of P is `(-2)/7, (-20)/7` Answer

Question (9) Find the coordinates of the points which divide the line segment joining A(–2, 2) and B(2, 8) into four equal parts.

Solution

Given, A(–2, 2) and B(2, 8)

Thus, coordinates of points which divide the given line segment in equal four parts.

Let, points that divide the given line segments in four equal parts are P, Q and R

We know that,

According to section formula, if the coordinates of the point P(x, y) divides the line segment joining the points A(x₁, y₁) and B(x₂, y₂), internally, in the ratio m₁:m₂ , then

`x=(m_1x_2+m_2x_1)/(m_1+m_2)` and `y=(m_1y_2+m_2y_1)/(m_1+m_2)`

Calculation of coordinate of point P

Here, x₁ = –2, x₂ = 2

And, y₁ = 2, y₂ = 8

And, ratio of in which point P divides the line = 1:3

Let the coordinate of point P is (x,y)

Thus, using section formula of coordinate geometry,

`x=((3xx(-2))+(1xx2))/(1+3)`

`=(-6+2)/4`

`=>x=(-4)/4 = -1`

And, `y=((1xx8)+(3xx2))/(1+3)`

`=(8+6)/4`

`=14/4`

`=>y=7/2`

Thus, coordinate of P is `(-1, 7/2)`

Calculation of coordinate of point Q, which divides the line in 2:2 ratio

Let, coordinate of point Q = (q₁, q₂)

Since, Q divides the line segment in the middle

Thus, coordinate of Q `=(-2+2)/2 , (2+8)/2`

`=0/2, 10/2`

= 0, 5

Thus, coordinate of Q is (0, 5)

Calculation of coordinate of point R which divides the line in the 3:1 ratio

Here, x₁ = –2, x₂ = 2

And, y₁ = 2, y₂ = 8

And, ratio of in which point R divides the line = 3:1

Let the coordinate of point R is (c, d)

Thus, using section formula of coordinate geometry

`c = ((3xx2)+(1xx(-2)))/(3+1)`

`=(6+(-2))/4`

`=(6-2)/4 = 4/4`

Thus, c = 1

And, `d=((3xx8)+(1xx2))/(3+1)`

`=(24+2)/4`

`=26/4 = 13/2`

Thus, coordinate of R is `(1, 13/2)`

Thus, required coordinates are `(-1, 7/2)`, (0, 5) and `(1, 13/2)` Answer

Question (10) Find the area of a rhombus if its vertices are (3, 0), (4, 5), (–1, 4) and (–2, –1) taken in order. [Hint: Area of rhombus = `1/2`(product of its diagonals)]

Solution

Given,

Vertices of rhombus are (3, 0), (4, 5), (–1, 4) and (–2, –1)

Thus, area of rhombus = ?

Let, the vertices of rhombus are A, B, C and D

Now, we know that, in coordinate geometry according to distance formula, in a plane the distance between two given points `=sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

Calculation of distance between A and C

Here, we have, x₁ = 3, x₂ = –1

And, y₁ = 0, y₂ = 4

Thus using distance formula,

AC `=sqrt((-1-3)^2+(4-0)^2)`

`=sqrt((-4)^2+(4)^2)`

`=sqrt(16+16)`

`=sqrt(32)`

`=sqrt(16xx2)`

⇒ AC `=4sqrt2`

Calculation of distance between B and D

Here, we have, x₁ = 4, x₂ = –2

And, y₁ = 5, y₂ = –1

Thus using distance formula,

BD `=sqrt((-2-4)^2+(-1-5)^2)`

`=sqrt((-6)^2+(-6)^2)`

`=sqrt(36+36)`

`=sqrt(72)`

`=sqrt(36xx2)`

⇒ BD`=6sqrt2`

Calculation of Area of Rhombus

Area of rhombus =`1/2`(product of diagonals)

`=1/2xx4sqrt2xx6sqrt2`

`=1/2xx24x2`

Thus, area of rhombus = 24 square unit Answer

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