R.S. Agarwal Solution Class 10 Co-ordinate Geometry Exercise 1B

Q1. Find the coordinate of the point which divide the line joining the points A (-1,7) and B(4, -3) in the ratio 2 : 3.

Let point P (X,Y) be the point divide the line joining points AB in 2 : 3
Here,

`x_1 = – 1`     `y_1 = 7`
`x_2 = 4`      `y_2 = – 3`     
`m_1 = 2`      `m_2= 3`
X = `frac{m_2 x_1 + m_1 x_2}{m_1 + m_2 }` = `frac{2(4) + 3(-1)}{2 + 3 }` = `frac{8- 3}{5 }` = 1

Y = `frac{m_2 y_1 + m_1 y_2}{m_1 + m_2 }` = `frac{2(-3) + 3(7)}{2 + 3 }` = `frac{-6 + 21}{5 }` = 3

Hence required point P is (1,3)

Q2. Find the coordinate of the point which divide the line joining the points P (-5, 11) and Q(4, -7) in the ratio 7 : 2.

Let point P (X,Y) be the point divide the line joining points AB in 7 : 2
Here,

`x_1 = – 5`     `y_1 = 11`
`x_2 = 4`      `y_2 = -7`     
`m_1 = 7`      `m_2= 2`
X = `frac{m_2 x_1 + m_1 x_2}{m_1 + m_2 }` = `frac{7(4) + 2(-5)}{7 + 2}` = `frac{28-10}{9 }` = `frac{18}{9}` =2

Y = `frac{m_2 y_1 + m_1 y_2}{m_1 + m_2 }` = `frac{7(-7) + 2(11)}{7 + 2 }` = `frac{-49 + 22}{9 }` = `frac{-27}{9}`= -3

Hence required point P is (2,-3)

Q3. Find the coordinates of the points which trisect of the line segment AB. whosh end points are A(2,1) and B(5, -8)

Let point P and Q be the points trisect the line joining AB.
Then P should be the point divide AB in 1: 2 and Q be the point divide AB in 2 : 1.

Now, for the point P

`x_1 = 2`     `y_1 = 1`
`x_2 = 5`      `y_2 = -8`     
`m_1 = 1`      `m_2= 2`
X = `frac{m_2 x_1 + m_1 x_2}{m_1 + m_2 }` = `frac{1(5) + 2(2)}{1 + 2}` = `frac{5+4}{3 }` = `frac{9}{3}` =3

Y = `frac{m_2 y_1 + m_1 y_2}{m_1 + m_2 }` = `frac{1(-8) + 2(1)}{1 + 2 }` = `frac{-8 + 2}{3 }` = `frac{-6}{3}`= -2

Hence required point P is (3,-2)

and for the point Q,

`x_1 = 2`     `y_1 = 1`
`x_2 = 5`      `y_2 = -8`     
`m_1 = 2`      `m_2= 1`
X = `frac{m_2 x_1 + m_1 x_2}{m_1 + m_2 }` = `frac{2(5) + 1(2)}{2 + 1}` = `frac{10+2}{3 }` = `frac{12}{3}` =4

Y = `frac{m_2 y_1 + m_1 y_2}{m_1 + m_2 }` = `frac{2(-8) + 1(1)}{2 + 1 }` = `frac{-16 + 1}{3 }` = `frac{-15}{3}`= -5

Hence required point P is (4,-5)

Q4. Find the coordinate of the points which divide the line joining the points A (-4,0) and B(0, 6) in three equal parts ?

Let point P and Q be the points trisect the line joining AB.
Then P should be the point divide AB in 1: 2 and Q be the point divide AB in 2 : 1.

Now, for the point P

`x_1 = -4`     `y_1 = 0`
`x_2 = 0`      `y_2 = 6`     
`m_1 = 1`      `m_2= 2`
X = `frac{m_2 x_1 + m_1 x_2}{m_1 + m_2 }` = `frac{1(0) + 2(-4)}{1 + 2}` = `frac{0-8}{3 }` = `frac{-8}{3}`

Y = `frac{m_2 y_1 + m_1 y_2}{m_1 + m_2 }` = `frac{1(6) + 2(0)}{1 + 2 }` = `frac{6 + 0}{3 }` = `frac{6}{3}` = 2

Hence required point P is (`frac{-8}{3}`,2)

and for the point Q,

`x_1 = -4`     `y_1 = 0`
`x_2 = 0`      `y_2 = 6`     
`m_1 = 2`      `m_2= 1`
X = `frac{m_2 x_1 + m_1 x_2}{m_1 + m_2 }` = `frac{2(0) + 1(-4)}{2 + 1}` = `frac{0-4}{3 }` = `frac{-4}{3}`

Y = `frac{m_2 y_1 + m_1 y_2}{m_1 + m_2 }` = `frac{2(6) + 1(0)}{2 + 1 }` = `frac{12 + 1}{3 }` = 4

Hence required point P is (`frac{-4}{3}`,4)

Q5. The line joining the points A (2,1) and B (5, -8) is trisected at the points P and Q. If the point P lies on the line 2x – y + k = 0. Find the value of k.

Let point P and Q be the points trisect the line joining AB.
Then P should be the point divide AB in 1: 2

Now, for the point P

`x_1 = 2`     `y_1 = 1`
`x_2 = 5`      `y_2 = -8`     
`m_1 = 1`      `m_2= 2`
X = `frac{m_2 x_1 + m_1 x_2}{m_1 + m_2 }` = `frac{1(5) + 2(2)}{1 + 2}` = `frac{5+4}{3 }` = `frac{9}{3}` =3

Y = `frac{m_2 y_1 + m_1 y_2}{m_1 + m_2 }` = `frac{1(-8) + 2(1)}{1 + 2 }` = `frac{-8 + 2}{3 }` = `frac{-6}{3}`= -2

Hence required point P is (3,-2)

Since point P lies on the line 2x – y + k = 0. We are putting the value of P in given equation.

2 (3) – (-2) + k = 0
6 + 2 + k = 0
k = -8

Q6. Let A (-1 , 5) and B (6, -2) be the two points. find the coordinates of the points P such that AP = `frac{3}{7}`AB.

Given that, AP = `frac{3}{7}`AB
⇒ `frac{7}{3}`=`frac{AB}{AP}`
⇒ `frac{7}{3}`=`frac{AP + PB}{AP}` ⇒ `frac{7}{3}`=`frac{AP + PB}{AP}` ⇒ `frac{7}{3}`=`frac{AP}{AP}`+ `frac{PB}{AP}`
⇒`frac{7}{3}` – 1 = `frac{PB}{AP}` ⇒ `frac{7-3}{3}` = `frac{PB}{AP}` ⇒ `frac{4}{3}` = `frac{PB}{AP}`
or `frac{AP}{PB}` = `frac{3}{4}`

Hence point P (X,Y) be the point divide the line joining points AB in 3 : 4
Here,

`x_1 = – 1`     `y_1 = 5`
`x_2 = 6`      `y_2 = – 2`     
`m_1 = 3`      `m_2= 4`
X = `frac{m_2 x_1 + m_1 x_2}{m_1 + m_2 }` = `frac{3(6) + 4(-1)}{3 + 4 }` = `frac{18- 4}{7 }` = `frac{14}{7 }`=2

Y = `frac{m_2 y_1 + m_1 y_2}{m_1 + m_2 }` = `frac{3(-2) + 4(5)}{3+4}` = `frac{-6 + 20}{7 }`=`frac{14}{7 }` = 2

Hence required point P is (2,2)

Q7.Find the coordinates of the midpoint of the line segmet joining:
(i)A(3,0) and B (5,4)
(ii) P(-11,-8) and Q(8,-2).

(i)

Let mid-point of AB is P(X,Y)

Using mid-point formula
X = `frac{x_1 + x_2}{2}`= `frac{3 + 5}{2}`= `frac{8}{2}` = 4
Y = `frac{y_1 + y_2}{2}`= `frac{0 + 4}{2}`= `frac{4}{2}` = 2
Hence mid pint of AB is P(4,2)
(ii)

Let mid-point of PQ is R(X,Y)

Using mid-point formula
X = `frac{x_1 + x_2}{2}`= `frac{-11 + 8}{2}`= `frac{-3}{2}`
Y = `frac{y_1 + y_2}{2}`= `frac{-8 -2}{2}`= `frac{-10}{2}` = -5
Hence mid pint of PQ is R(`frac{-3}{2}`,-5)

Q8. Find the coordinates of the midpoint of the line segment joining the points A(2p + 1, 4) and B(5,q-1) are (2p,q). Find the value of p and q

Let P (X,Y) be the mid-point of AB.

Then P = (`frac{x_1 + x_2}{2}`, `frac{y_1 + y_2}{2}`) = (`frac{2p + 1 + 5}{2}`, `frac{4 + q-1}{2}`)
⇒(`frac{2p + 6}{2}`, `frac{q + 3}{2}`) =(2p,q) (∵ mid point of AB is (2p, q))
Equating both sides, `frac{2p + 6}{2}` = 2p and `frac{q + 3}{2}`= q
⇒ 2p + 6 = 4p and q + 3 = 2q
⇒ 6 = 4p – 2p and 3 = 2q -q
⇒ 6 = 2p and 3 = q
⇒ p = 3 and q = 3

Q9. Find the coordinates of the midpoint of the line segment joining the points A(2a, 4) and B(-2,3b) are M(1,2a + 1). Find the value of a and b.

Let M (X,Y) be the mid-point of AB.

Then M = (`frac{x_1 + x_2}{2}`, `frac{y_1 + y_2}{2}`) = (`frac{2a – 2}{2}`, `frac{4 + 3b}{2}`)
⇒(a -1, `frac{4 + 3b}{2}`) = (1,2a + 1) (∵ mid point of AB is (1,2a + 1)) Equating both sides,
a -1 = 1     and     `frac{4 + 3b}{2}`= 2a + 1
⇒ a = 1 + 1     and     2a + 1 = `frac{4 + 3b}{2}`
⇒ a = 2     and     4a + 2 = 4 + 3b —(Equation 1)
Put value of a in equation (1)
4(2) + 2 = 4 + 3b⇒ 8 + 2 – 4 = 3b ⇒ 6 = 3b ⇒ b = 2

Q10. The line segment joining A (-2, 9) and B(6,3) is a diameter of a circle with centre C. Find the coordinate of C

Since C is the centre of the circle. C will be the mid point of the diametre AB

Hence by the using mid point formula. We get,
C (`frac{-2 + 6}{2}`, `frac{9 + 3}{2}`) = (`frac{4}{2}`, `frac{12}{2}`) = (2,6)

Q11. AB is a diameter of a circle with centre C (-1, 6). If the coordinates of A are (-7,3). Find the coordinates of B.

Let point B be (x, y)

Since C is mid point of AB ⇒ C (`frac{-7 + x}{2}`, `frac{3 + y}{2}`) = (- 1, 6) Equating both sides. We get,
⇒ `frac{-7 + x}{2}= – 1` and `frac{3 + y}{2}= 6`
⇒ `- 7 + x =-1` and `3 + y = 12`
⇒ `x = 6` and `y = 9`
Hence point B is (6, 9)

Q12. In what ratio does the point P(2,5) divide the join of A(8,2) and B(-6,9)?

Let point P (2,5) be the points divides AB in k : 1

Here,
`x_1 = 8`     `y_1 = 2`  
`x_2 = -6`      `y_2 = -9`     
`m_1 = k`      `m_2= 1`
X = `frac{m_1 x_2 + m_2 x_1}{m_1 + m_2 }` = `frac{k(-6) + 1(8)}{k + 1 }` = `frac{-6k + 8}{k + 1 }` = 2
⇒ 2 ( k + 1) = – 6k + 8
⇒ 2k + 2 = – 6 k + 8
⇒ 2k + 6k = 8 -2
⇒ 8k = 6
⇒ k = `frac{6}{8}` = `frac{3}{4}`
Hence, required ratio is 3 : 4

Q13. Find the ratio in which the point P (-6, a) divides A (-3, -1)and B (-8, 9). Also. find the value of a.

Let point P (-6,a) be the points divides AB in k : 1

Here,
`x_1 = -3`     `y_1 = -1`  
`x_2 = -8`      `y_2 = 9`     
`m_1 = k`      `m_2= 1`
X = `frac{m_1 x_2 + m_2 x_1}{m_1 + m_2 }` = `frac{k(-8) + 1(-3)}{k + 1 }` = `frac{-8k -3}{k + 1 }` = -6
⇒ -6 ( k + 1) = – 8k -3
⇒ -6k -6 = – 8 k -3
⇒ -6+ 8k = -3 + 6
⇒ 2k = 3
⇒ k = `frac{3}{2}` 
Hence, required ratio is 3 : 2
Now, `y_1 = 2`  
`y_2 = -9`     
`m_1 = k`      `m_2= 1` and Y= a
a = `frac{m_1 y_2 + m_2 y_1}{m_1 + m_2 }` = `frac{3(9) + 2(-1)}{3 + 2 }` = `frac{27 – 2}{5 }` = 5

Q14. Find the ratio in which the point P (a,1) divides the join of A(-4,4) and B (6,-1) and hence find the value of a.

Let point P (-6,a) be the points divides AB in k : 1

Here,
`x_1 = -4`     `y_1 = 4`  
`x_2 = 6`      `y_2 = -1`     
`m_1 = k`      `m_2= 1`
Y = `frac{m_1 y_2 + m_2 y_1}{m_1 + m_2 }` = `frac{k(-1) + 1(4)}{k + 1 }` = `frac{-k + 4}{k + 1 }` = 1
⇒  ( k + 1) = – k + 4
⇒ k + 1 = – k  + 4
⇒ k+ k = 4 -1
⇒ 2k = 3
⇒ k = `frac{3}{2}` 
Hence, required ratio is 3 : 2
Now, `x_1 = -4`  
`x_2 = 6`     
`m_1 = 3`      `m_2= 2` and X= a
a = `frac{m_1 x_2 + m_2 x_1}{m_1 + m_2 }` = `frac{3(6) + 2(-4)}{3 + 2 }` = `frac{18 – 8}{5 }` =  2

Q15. In what ratio is the line segment joining A (2,-3) and B (5, 6) divided by the x -axis? Also, find the coordinates of the point of division.

Let point P (X,0) be the point on x-axis divides AB in k : 1

Here,
`y_1 = 4`  
`y_2 = -1` and Y = 0
`m_1 = k`      `m_2= 1`
Y = `frac{m_1 y_2 + m_2 y_1}{m_1 + m_2 }` = `frac{k(6) + 1(-3)}{k + 1 }` = `frac{6k -3 }{k + 1 }` = 0
⇒0 ( k + 1) = 6k -3
⇒ 0 = 6k -3
⇒ 6k = 3
⇒ k = `frac{3}{6}` or `frac{1}{2}`
Hence, required ratio is 1 : 2
Now, to find the coordinate P (X, 0). We have
`x_1 = 2`
`x_2 = 5`
`m_1 = 1`      `m_2= 2`
X = `frac{m_1 x_2 + m_2 x_1}{m_1 + m_2 }` = `frac{1(5) + 2(2)}{1 + 2 }` = `frac{5 + 4}{3 }` = 3
Hence point P is (3,0 )

Q16. In what ratio is the line segment joining the points A(-2, -3) and B (3,7) divided by the y -axis? Also, find coordinates of the point of division

Let point P (0,Y) be the point on y-axis divides AB in k : 1

Here,
`x_1 = -2`
`x_2 = 3` and X = 0
`m_1 = k`      `m_2= 1`
X = `frac{m_1 x_2 + m_2 x_1}{m_1 + m_2 }` = `frac{k(3) + 1(-2)}{k + 1 }` = `frac{3k -2 }{k + 1 }` = 0
⇒0 ( k + 1) = 3k -2
⇒ 0 = 3k -2
⇒ 3k = 2
⇒ k = `frac{2}{3}`
Hence, required ratio is 2 : 3
Now, to find the coordinate P (0, Y). We have
`y_1 = -3`
`y_2 = 7`
`m_1 = 2`      `m_2= 3`
Y = `frac{m_1 y_2 + m_2 y_1}{m_1 + m_2 }` = `frac{2(7) + 3(-3)}{2 + 3 }` = `frac{14 -9}{5 }` = 1
Hence point P is (0,1 )

Q17. If A(5,-1), B(-3,-2) and C (-1,8) are the vertices of △ABC. Find the length of the median through A and the coordinates of the centroid.

To find the length of the median AD. Firstly, we have to find point D, which is mid point of BC by using mid point formula.
D = (`frac{x_1 + x_2}{2}`, `frac{y_1 + y_2}{2}`) = (`frac{-3 -1}{2}`, `frac{-2 + 8}{2}`)= (-2, 3)
Now length of median
AD = `sqrt{(x_2 – x_ 1)^ 2+(y_2 – y_ 1)^ 2}`
AD = `sqrt{(-2 – 5)^ 2+(3 – (-1))^ 2}`
AD = `sqrt{(-7)^ 2+(4)^ 2}`
AD = `sqrt{65}`
Let centroid o divides AD in 2 : 1
Here,
`x_1 = 5`
`x_2 = -2` and `y_1 = -1`
`y_2 = 3`
`m_1 = 2`      `m_2= 1`
X = `frac{m_1 x_2 + m_2 x_1}{m_1 + m_2 }` = `frac{2(-2) + 1(5)}{2 + 1 }` = `frac{-4 + 5}{3}` = 1/3
Y = `frac{m_1 y_2 + m_2 y_1}{m_1 + m_2 }` = `frac{2(3) + 1(-1)}{2 + 1 }` = `frac{6 -1}{3 }` = `frac{5}{3 }`

Q18. Find the lengths of the medians of a △ABC whose vertices are A(0, -1), B(2, 1) and C (0,3)

To find the length of the median AD. Firstly, we have to find point D, which is mid point of BC, by using mid point formula.
D = (`frac{x_1 + x_2}{2}`, `frac{y_1 + y_2}{2}`) = (`frac{2 + 0}{2}`, `frac{1 + 3}{2}`)= (1, 2)
Now length of median
AD = `sqrt{(x_2 – x_ 1)^ 2+(y_2 – y_ 1)^ 2}`
AD = `sqrt{(1 -0 )^ 2+(2 + 1)^ 2}`
AD = `sqrt{ 1 + 9}`
AD = `sqrt{10}`

To find the length of the median BE. Firstly, we have to find point E, which is mid point of AC, by using mid point formula.
D = (`frac{x_1 + x_2}{2}`, `frac{y_1 + y_2}{2}`) = (`frac{0 + 0}{2}`, `frac{- 1 + 3}{2}`)= (0, 1)
Now length of median
AD = `sqrt{(x_2 – x_ 1)^ 2+(y_2 – y_ 1)^ 2}`
AD = `sqrt{(0 -2 )^ 2+( 1 -1)^ 2}`
AD = `sqrt{ 4 + 0}`
AD = `2`

To find the length of the median CF. Firstly, we have to find point F, which is mid point of BA, by using mid point formula.
D = (`frac{x_1 + x_2}{2}`, `frac{y_1 + y_2}{2}`) = (`frac{2 + 0}{2}`, `frac{1 -1}{2}`)= (1 , 0)
Now length of median
AD = `sqrt{(x_2 – x_ 1)^ 2+(y_2 – y_ 1)^ 2}`
AD = `sqrt{(1 – 0)^ 2+(0 -3)^ 2}`
AD = `sqrt{1  + 9}`
AD = `sqrt{10}`

Q19. Find the coordinate of the centroid of △ABC. whose vertices are A(6, -2), B (4, -3) and C (-1,-4)..

Method 1

We know that, centroid divides median in 2: 1. Therefore, to find centroid we need point D (mid point of BC).
D = (`frac{x_1 + x_2}{2}`, `frac{y_1 + y_2}{2}`) = (`frac{4 – 1}{2}`, `frac{-3 – 4}{2}`)= (3/2 , -7/2)
Let point G (X,Y) is centroid which divides median AD 2 : 1
Here,

`x_1 = – 6`     `y_1 = -2`
`x_2 = 3/2`      `y_2 = -7/2`     
`m_1 = 2`      `m_2= 1`
X = `frac{m_2 x_1 + m_1 x_2}{m_1 + m_2 }` = `frac{2(3/2) + 1(6)}{2 + 1}` = `frac{3 + 6}{3}` = `frac{9}{3}` =3

Y = `frac{m_2 y_1 + m_1 y_2}{m_1 + m_2 }` = `frac{2(-7/2) + 1(-2)}{2 + 1 }` = `frac{-9}{3 }` = -3

Hence required point G is (3,-3)

Method 2

By using centroid formula G = (`frac{x_1 + x_2 + x_3}{3}`, `frac{y_1 + y_2 + y_3}{3}`) = (`frac{6 + 4 – 1}{3}`, `frac{-2 -3 -4}{3}`)= (3 , -3)

Q20. Find the centroid of △ABC whose vertices are A(-1,0), B(5,-2) and C(8,2).

Method 1

We know that, centroid divides median in 2: 1. Therefore, to find centroid we need point D (mid point of BC).
D = (`frac{x_1 + x_2}{2}`, `frac{y_1 + y_2}{2}`) = (`frac{5 + 8}{2}`, `frac{-2 + 2}{2}`)= (13/2 , 0)
Let point G (X,Y) is centroid which divides median AD 2 : 1
Here,

`x_1 = – 1`     `y_1 = 0`
`x_2 = 13/2`      `y_2 = 0`     
`m_1 = 2`      `m_2= 1`
X = `frac{m_2 x_1 + m_1 x_2}{m_1 + m_2 }` = `frac{2(13/2) + 1(-1)}{2 + 1}` = `frac{12}{3}` =4

Y = `frac{m_2 y_1 + m_1 y_2}{m_1 + m_2 }` = `frac{2(0) + 1(0)}{2 + 1 }` = `frac{0}{3 }` = 0

Hence required point G is (4,0)

Method 2

By using centroid formula G = (`frac{x_1 + x_2 + x_3}{3}`, `frac{y_1 + y_2 + y_3}{3}`) = (`frac{-1 + 5 + 8}{3}`, `frac{0 – 2 + 2}{3}`)= (4 , 0)

Q21. A (3,2) and B(-2,1) are two vertices of a △ABC, whose centroid is G (5/3 , -1/3). Find the coordinates of the third vertex C.

Let third vertix is (x,y)
By using centroid formula
G = (`frac{x_1 + x_2 + x_3}{3}`, `frac{y_1 + y_2 + y_3}{3}`) = (`frac{3- 2 + x}{3}`, `frac{2 + 1 + y}{3}`)
= (`frac{1+ x}{3}`, `frac{ 3+ y}{3}`) = (5/3 , -1/3) (Given centroid G = (5/3 , -1/3))
Equating both sides
`frac{1+ x}{3}`= `frac{5}{3}` and `frac{ 3+ y}{3}` = `frac{-1}{3}`
or 1 + x = 5 and 3 + y = -1
or x = 4 and y = – 4
Hence third vertex is (4, -4)

Q22. If G (-2, 1) is the centroid of a △ABC and two of its vertices are A(1,-6) and B(-5,2). Find the third vertex of the triangle.

Let third vertix is (x,y)
By using centroid formula
G = (`frac{x_1 + x_2 + x_3}{3}`, `frac{y_1 + y_2 + y_3}{3}`) = (`frac{1 -5 + x}{3}`, `frac{-6 + 2 + y}{3}`)
= (`frac{- 4+ x}{3}`, `frac{ -4 + y}{3}`) = (- 2, 1) (Given centroid G = (- 2, 1) )
Equating both sides
`frac{- 4+ x}{3}`= – 2 and `frac{ -4 + y}{3}` = 1
or -4 + x = – 6 and – 4 + y = 3
or x = – 6 + 4 and y = 3 + 4
or x = – 2 and y = 7
Hence third vertex is (- 2, 7)

Q23. Find the third vertex of a △ABC if two of its vertices are B (-3, 1) and C (0, -2) and its centroid is at the origin.

Let vert A ix is (x,y)
By using centroid formula
G = (`frac{x_1 + x_2 + x_3}{3}`, `frac{y_1 + y_2 + y_3}{3}`) = (`frac{x -3 + 0}{3}`, `frac{y + 1- 2}{3}`)
= (`frac{x- 3}{3}`, `frac{ y – 1}{3}`) = (0, 0) (Given centroid G is origin )
Equating both sides
`frac{ x – 3}{3}`= 0 and `frac{y – 1}{3}` = 0
or x- 3 = 0 and y – 1 = 0
or x = 3 and y = 1
Hence third vertex is (3, 1)

Q24. Show the the points A (3,1), B (0, -2 ), C (1,1) and D (4,4) are the vertices of parellelogram ABCD.

Using distance formula
AB = `sqrt{(x_2 – x_ 1)^ 2+(y_2 – y_ 1)^ 2}`
= `sqrt{(0 -3)^ 2+(-2 -1)^ 2}`
= `sqrt{9 + 9}`
= `sqrt{(18}`
= 3`sqrt{(2}`
CD = `sqrt{(x_2 – x_ 1)^ 2+(y_2 – y_ 1)^ 2}`
= `sqrt{(4-1)^ 2+(4-1)^ 2}`
= `sqrt{9 + 9}`
= `sqrt{(18}`
= 3`sqrt{(2}`
AD = `sqrt{(x_2 – x_ 1)^ 2+(y_2 – y_ 1)^ 2}`
= `sqrt{(4 -3)^ 2+(4-1)^ 2}`
= `sqrt{1 + 9}`
= `sqrt{(10}`
BC = `sqrt{(x_2 – x_ 1)^ 2+(y_2 – y_ 1)^ 2}`
= `sqrt{(1 -0)^ 2+(-1 – 3)^ 2}`
= `sqrt{1 + 9}`
= `sqrt{(10}`
Since pair of opposite sides are equal. Hence ABCD are vertices of ||grm.

Q25. It the points P (a, -11), Q (5,b), R (2.15) and S (1.1) are the vertices of a parallelogram PQES. Find the value of a and b.

O is mid point of PR
O = (`frac {a + 2}{2}`, `frac {- 11 + 15}{2}`)
= (`frac {a + 2}{2}`, `frac { 4}{2}`)
= (`frac {a + 2}{2}`, 2)
and also, O is mid point of QS
O = (`frac {5 + 1}{2}`, `frac {b + 1}{2}`) = (3, `frac {b + 1}{2}`)
Since Diagonals of ||grm bisecting each other at O
(`frac {a + 2}{2}`, 2) = (3, `frac {b + 1}{2}`)
Equating both sides.
`frac {a + 2}{2}` = 3 and 2 = `frac {b + 1}{2}`
a + 2 = 6 and b + 1 = 4
a = 4 and b = 3

Q26.If three consecutive vertices of a parallelogram ABCD are A (1,-2), B (3,6) and C (5,10), Find its fourth vertices D.

O is mid point of PR
O = (`frac {1 + 5}{2}`, `frac { – 2 + 10}{2}`)
= (`frac {6}{2}`, `frac { 8}{2}`)
= (3, 4)
and also, O is mid point of QS
O = (`frac { 3 + x}{2}`, `frac { 6 + y}{2}`)
Since Diagonals of || grm bisecting each other at O
(`frac { 3 + x}{2}`, `frac { 6 + y}{2}`) = (3, 4)
Equating both sides.
`frac { 3 + x}{2}` = 3 and `frac { 6 + y}{2}`= 4
3 + x = 6 and 6 + y = 8
x = 3 and y = 2
Hence fourth vertices is (3,2)