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.
Solution:

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.
Solution:

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)
Solution:

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 ?
Solution:

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.
Solution:

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.
Solution:

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).
Solution:

(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
Solution:

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.
Solution:

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
Solution:

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.
Solution:

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)?
Solution:

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.
Solution:

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.
Solution:

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.
Solution:

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
Solution:

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.
Solution:

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)
Solution:

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)..
Solution:

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).
Solution:

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.
Solution:

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.
Solution:

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.
Solution:

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.
Solution:

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.
Solution:

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.
Solution:

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)

Leave a Comment

Your email address will not be published. Required fields are marked *

Scroll to Top