Coordinate Geometry

Mathematics Class Tenth

10th-Math-home


NCERT Exercise 7.4(optional)

Question (1) Determine the ratio in which the line 2x + y – 4 =0 divides the line segment joining the points A(2, –2) and B(3, 7).

Solution :

Given points joining the line segment are A(2, –2) and B(3, 7)

Dividing line 2x + y – 4 =0 - - - - -(i)

Thus, ratio of division = ?

Let, the given line divides the line segment joining the points A(2, –2) and B(3, 7) in the k:1 ratio.

Now, we know that, according to section formula when a point P(x,y) divides the line segment joining the points Q(x1, y1) and R(x2, y2) in the k:1 ratio, then

P(x, y) `=((kx_2+x_1)/(k+1), (ky_2+y_1)/(k+1))`

Here we have, x1 = 2, y1 = –2

And, x2 = 3, y2 = 7

Thus, using section formula

`x=((kxx3+2)/(k+1))`

`=>x=((3k+2)/(k+1))`

And, `y=((kxx7+(-2))/(k+1))`

`=>y=((7k-2)/(k+1))`

Now, We have the given line in question

2x + y – 4 = 0

Now, after substituting the values of x and y in the given line in question, we get

`2((3k+2)/(k+1))+((7k-2)/(k+1))-4=0`

`=>(6k+4)/(k+1)+ (7k-2)/(k+1)-4=0`

`=>((6k+4+7k-2)-4(k+1))/(k+1)=0`

After cross multiplication, we get

6 k + 4 + 7 k – 2 –4(k + 1)=0

⇒ 6k + 4 + 7 k – 2 – 4k – 4 =0

After rearranging the above equation

⇒ 6 k + 7 k – 4 k –2 –4 + 4 = 0

⇒ 9 k – 2 = 0

⇒ 9 k = 2

`=> k = 9/2`

`=>k/1 = 9/2`

Thus, k : 1 = 9 : 2

Thus, required ratio is 9 : 2 Answer

Question (2) Find the relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.

Solution

Given points are (x, y), (1, 2) and (7, 0), which are collinear.

Then value of x and y = ?

Let, the points form a triangle.

And as given in question, points are collinear, thus area of triangle will be zero, because points are collinear.

We know that, if triangle formed by three points (x1, y1), (x2, y2), and (x3, y3),

Then, area of triangle

= 1/2 [x1 (y2 – y2) + x2 (y3 – y1) + x3 (y1 – y2)]

Here, we have, x1 = x, y1 = y

And, x2 = 1, y2 = 2

And, x3 = 7, y3 = 0

Thus Area of given triangle

= 1/2 [x (2 – 0) + 1 (0 – y) + 7 (y – 2)]

= 1/2 [{x × 2} + {1 (–y)} + {7y – 14} ]

= 1/2 [–2 x – y + 7 y – 14]

= 1/2 [2x + 6 y – 14]

= 1/2 × 2 [x + 3 y – 7]

= x + 3 y – 7

Now, since area of triangle is equal to zero, because points are collinear

Thus, x + 3 y – 7 = 0

⇒ x + 3y = 7

Thus, relation between x and y is x + 3 y = 7 Answer

Question (3) Find the centre of a circle passing through the points (6, –6), (3, –7) and (3, 3)

Solution

Given, points of circle are (6, –6), (3, –7) and (3, 3)

Then, coordinate of centre of given circle = ?

Let, the points of circle are A(6, –6), B(3, –7) and C(3, 3)

And centre of circle = O(x, y)

10 math coordinate geometry ncert exercise 7.4 question3

Now, since OA, OC and OB are radii of same circle,

Thus, OA = OB = OC

Now, we know that, in coordinate geometry, the distance between two points P (x1, y1) and Q (x2, y2), i.e

PQ `=sqrt((x_1-x_2)^2+(y_1-y_2)^2)`

Calculation of OA

Here we have, x1 = x, y1 = y

And, x2 = 6, y2 = –6

Thus, accordinat to distance formula of coordinate geometry

OA `=sqrt((6-x)^2+(y-(-6))^2)`

⇒ OA `=sqrt((6-x)^2+(y+6)^2)` - - - - - (i)

Calculation of OB

Here we have, x1 = x, y1 = y

And, x2 = 3, y2 = –7

Thus, accordinat to distance formula of coordinate geometry

OB `=sqrt((x-3)^2+(y-(-7))^2)`

⇒ OB `=sqrt((x-3)^2+(y+7)^2)` - - - - - (ii)

Calculation of OC

Here we have, x1 = x, y1 = y

And, x2 = 3, y2 = 3

Thus, according to distance formula of coordinate geometry

OC `=sqrt((x-3)^2+(y-3)^2)` - - - - (iii)

Now, since, OA = OB [since they are radii of same circle]

Thus, from equation (i) and (ii) we have

`sqrt((6-x)^2+(y+6)^2)` `=sqrt((x-3)^2+(y+7)^2)`

After squaring both sides, we get

(6 – x)2 + (y + 6)2 = (x – 3)2 + (y + 7)2

⇒ 62 + x2 – 2×6×x + y2 + 62 + 2 × y × 6 = x2 + 32 – 2 × x × 3 + y2 + 72 + 2 × y × 7

⇒ 36 + x2 – 12x + y2 + 36 + 12 y = x2 + 9 – 6x + y2 + 49 + 14 y

After rearranging the above equation

⇒ x2 + y2 – 12 x + 12 y + 36 + 36 = x2 + y2 – 6 x + 14 y + 9 + 49

After transposing the RHS to LHS, we get

⇒ x2 + y2 – 12 x + 12 y + 36 + 36 – (x2 + y2 – 6 x + 14 y + 9 + 49) = 0

x2 + y2 – 12 x + 12 y + 36 + 36 – x2y2 + 6x – 14 y – 58 = 0

⇒ – 12 x + 12 y + 6x – 14 y + 72 – 58 = 0

⇒ – 6x – 2y + 14 = 0 - - - - (iv)

Now, since, OB = OC [Because both are radii of same circle]

Thus, from equation (ii) and (iii), we get

`sqrt((x–3)^2+(y+7)^2)` `=sqrt((x–3)^2+(y–3)^2)`

⇒ x2 + 32 – 2 × x × 3 + y2 72 + 2 × y × 7 = x2 + 32 – 2 × x × 3 + y2 + 32 – 2 × y × 3

x2 + 9 – 6 x + y2 + 49 + 14 y = x2 + 9 – 6x + y2 + 9 – 6 y

⇒ 9 – 6 x + 49 + 14 y = 9 – 6x + 9 – 6 y

⇒ 9 + 49 +14 y = 9 + 9 – 6 y

⇒ 58 + 14 y = 18 – 6 y

⇒ 14 y + 6 y = 18 – 58

⇒ 20 y = – 40

`=>y = (–40)/20 = –2`

Thus, y = – 2

Now, from equation (iv), we have

– 6 x – 2y + 14 = 0

Now, after substituting the value of y in the equation (iv) as given above, we get

– 6 x – 2 (– 2) + 14 = 0

⇒ – 6 x + 4 + 14 = 0

⇒ – 6 x + 18 = 0

⇒ 18 = 6 x

`=> x = 18/6 =3`

Thus, x = 3

Thus, coordinate of centre of given circle is (3, –2) Answer

Question (4) The two opposite vertices of a square are (–1, 2) and (3, 2). Find the coordinates of the other two vertices.

Solution

Given, two opposite vertices of a square = (–1, 2) and (3, 2)

Then, other two vertices = ?

Let, ABCD is a square and whose opposite vertices A(–1, 2) and C(3, 2) are given

10 math coordinate geometry ncert exercise 7.4 question4

Now, we know that, in coordinate geometry, the distance between two points P (x1, y1) and Q (x2, y2), i.e

PQ `=sqrt((x_1-x_2)^2+(y_1-y_2)^2)`

Calculation of distance between C and D i.e. CD

Let coordinates of point D = x, y

Here we have, x1 = 3, y1 = 2

And, x1 = x, y1 = y

Thus, according to distance formula of coordinate geometry

CD `=sqrt((3-x)^2 + (2-y)^2)` - - - - - - (i)

Calculation of distance between A and D, i.e. AD

Here we have, x1 = –1, y1 = 2

And, x1 = x, y1 = y

Thus, according to distance formula of coordinate geometry

AD `=sqrt((-1-x)^2+(2-y)^2)` - - - - - (ii)

Now, since CD and Ad are side of same square,

Thus, CD = AD

Thus, from equation (i) and (ii), we have

`sqrt((3-x)^2 + (2-y)^2)` `=sqrt((-1-x)^2+(2-y)^2)`

After squaring both sides, we get

(3 – x)2 + (2 – y)2 = (– 1 – x)2 + (2 – y)2

⇒ 32 + x2 – 2 × 3 × x + 22 + y2 – 2 × 2 × y = (– 1)2 + x2 – 2 (– 1) x + 22 + y2 – 2 × 2 × y

⇒ 9 + x2 – 6 x + 4 + y2 – 4 y = 1 + x2 + 2x + 4 + y2 – 4 y

⇒ 9 – 6 x + 4 – 4 y = 1 + 2 x + 4 – 4 y

⇒ 13 – 6 x = 5 + 2 x

⇒ – 6 x – 2 x = 5 – 13

⇒ – 8 x = – 8

⇒ 8 x = 8

`=>x = 8/8 =1`

Thus, x = 1

Calculation of distance between A and C, i.e. AC

Here we have, x1 = –1, y1 = 2

And, x1 = 3, y1 = 2

Thus, according to distance formula of coordinate geometry

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

`=sqrt((-4)^2+0^2))`

`=sqrt(16)`

⇒ AC = 4 - - - - (iii)

Now, In right angle triangle ACB

AC2 = AD2 + DC2

After substituting the value from of AC, AD and CD from equation (iii), (ii) and (i), we get

42 `= (sqrt((-1-x)^2+(2-y)^2))^2` `+ (sqrt((3-x)^2 + (2-y)^2))^2`

⇒ 16 = {(– 1 – x)2 + (2 – y)2} + {(3 – x)2 + (2 – y)2}

⇒ 16 = {(– 1)2 + x2 – 2 (– 1) x + 22 + y2 – 2 × 2 × y} + {32 + x2 – 2 × 3 × x + 22 + y2 – 2 × 2 × y }

⇒ 16 = {1 + x2 + 2x + 4 + y2 – 4 y } + {9 + x2 – 6 x + 4 + y2 – 4 y }

⇒ 16 = 1 + x2 + 2x + 4 + y2 – 4 y + 9 + x2 – 6 x + 4 + y2 – 4 y

After rearranging the above expression

⇒ 16 = x2 + x2 + y2 + y2 + 2 x – 6 x – 4 y – 4 y + 1 + 4 + 9 + 4

⇒ 16 = 2 x2 + 2 y2 – 4 x – 8 y + 18

Now, after substituting the value of x = 1, we get

16 = 2 (1)2 + 2 y2 – 4 × 1 – 8 y + 18

⇒ 16 = 2 + 2 y2 – 4 – 8 y + 18

⇒ 16 = 2 y2 – 8 y + 16

⇒ 2 y2 – 8 y = 16 – 16

⇒ 2 y2 – 8 y = 0

⇒ 2y (y – 4) = 0

Now, if 2y = 0

Therefore, y = 0

Or, if y – 4 = 0

Then, y = 4

Thus, y = 0 or 4

Now, we know that, diagonals of a square cut each other at middle point.

In the figure we assumed the O is the middle point of AC and DB

Calculation of middle point O of AC

We know that, coordinates of middle point joining the two points P(x1, y1) and Q(x2, y2) are

`((x_1+x_2)/2, (y_1+y_2)/2)`

Thus, coordinates of middle point of AC

Here, we have, x1 = –1, y1 = 2

And, x2 = 3, y2 = 2

Thus, coordinate of mid-point of AC, O are

`((-1+3)/2, (2+2)/2)`

`=(2/2, 4/2)`

⇒ Coordinates of O = (1, 2) - - - - (iv)

Calculation of mid-point O of BD

We have coordinate of D = (1, 0) or (1, 4)

Let, the coordinates of B = (x, y)

Thus, here we have, x1 = x, y1 = y

And, x2 = 1, y2 = 0 or 4

Thus, coordinates of mid-point O of BD

`=((x+1)/2, (y+y_1)/2)`

Since, coordinates of O = (1, 2) as calculated above [equation (iv)]

Thus, (1, 2) `=((x+1)/2, (y+y_1)/2)`

Thus, `(x+1)/2=1` and `(y+y_1)/2 = 2`

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

After cross multiplication, we get

x + 1 = 1 × 2

⇒ x + 1 = 2

Therefore, x = 2 – 1 = 1

Now, for `(y+y_1)/2 = 2`

When y1 = 0

Then, `(y+0)/2 = 2`

`=>y/2 = 2`

After cross multiplication, we get

y = 2 × 2 = 4

And when y1 = 4

Then `(y+y_1)/2 = 2`

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

After cross multiplication, we get

y + 4 = 2 × 2

⇒ y + 4 = 4

⇒ y = 4 – 4 = 0

Thus, y = 4 or 0

Thus, coordinates of D = (1, 4) or (1, 0)

Thus, coordinates of opposite vertices of given square are (1, 0) and (1, 4) Answer

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