ML Aggarwal Solution Class 10 Chapter 11 Section Formula Test

 Test

Question 1

The base BC of an equilateral triangle ABC lies on y-axis. The coordinates of the point C are (0, – 3). If origin is the mid-point of the base BC, find the coordinates of the points A and B

Sol :

Base BC of an equilateral ∆ABC lies on y-axis

co-ordinates of point C are (0, – 3),

origin (0, 0) is the mid-point of BC.

Figure to be added











Let co-ordinates of B be (x, y)

0=x+02x2=0x=0

y32=0y3=0

y=3

Co-ordinates of B are (0,3)

Again let co-ordinates of A be (x, 0) as it lies on x-axis.

AB=AC=BC=6 units

=(x0)2+(03)2=62

x2+(3)2=62

x2+9=36x2=369=27

x=±33

Co-ordiantes of A will be (±33,0)


Question 2

A and B have co-ordinates (4, 3) and (0, 1), Find

(i) the image A’ of A under reflection in the y – axis.

(ii) the image of B’ of B under reflection in the lineAA’.

(iii) the length of A’B’.

Sol :
Figure to be added










(i) Co-ordinates of A’, the image of A (4, 3)

reflected in y-axis will be ( – 4, 3).

(ii) Co-ordinates of B’ the image of B (0, 1)

reflected in the line AA’ will be (0, 5).

(iii) Length A’B’

=[0(4)]2+(53)2
=(4)2+(2)2=16+4
=20=4×5=25 units


Question 3

Find the co-ordinates of the point that divides the line segment joining the points P (5, – 2) and Q (9, 6) internally in the ratio of 3 : 1.

Sol :

Let R be the point whose co-ordinates are (x, y)

which divides PQ in the ratio of 3:1.

x=m1x2+m2x1m1+m2

=3×9+1×53+1=27+54=324=8

y=m1y2+m2y1m1+m2=3×6+1×(2)3+1

=1824=164=4

Co-ordinates of R will be (8,4)


Question 4

Find the coordinates of the point P which is three-fourth of the way from A (3, 1) to B ( – 2, 5).

Sol :

Co-ordinates of A (3, 1) and B ( – 2, 5)

P lies on AB such that

AP=34AB=34(AP+PB)

AP=3 PBAP:PB=3:1

Let co-ordinates of P be (x, y)

x=mx2+nx1m+n=3×(2)+1×33+1

=6+34=34


y=my2+ny1m+n=3×5+1×13+1

=15+14=164=4

Co-ordinates of P are (34,4)


Question 5

P and Q are the points on the line segment joining the points A (3, – 1) and B ( – 6, 5) such that AP = PQ = QB. Find the co-ordinates of P and Q.

Sol :

Given

AP = PQ = QB





P divides AB in the ratio of 1: 2 and Q divides it in 2: 1

Let co-ordinates of P will be (x1,y1) and of Q will be (x2,y2)

x1=m1x2+m2x1m1+m2=1×(6)+2(3)1+2

=6+63=03=0


y1=m1y2+m2y1m1+m2

=1×5+2(1)1+2=523=33=1

Co-ordinates of P will be (0,1)

Again

x2=m1x2+m2x1m1+m2=2×(6)+1(3)2+1

=12+33=93=3


y2=m1y2+m2y1m1+m2=2×5+1(1)2+1

=1013=93=3

Co-ordinates of Q will be (-3,3)


Question 6

The centre of a circle is (α + 2, α – 5). Find the value of a given that the circle passes through the points (2, – 2) and (8, – 2).

Sol :

Let A (2, -2), B (8, -2) and centre of the circle be

O (α + 2, α – 5)

OA=OB= radii of the same wil

OA=(2α2)2+(2α+1)2

=α2+(1α)2=α2+(α+1)2..(i)

OB=(8α2)2+(2α+1)2

=(6α)2+(1α)2

=(6α)2+(1+α)2...(ii)

From (i) and (ii)

α2+(α+1)2=(6α)2+(1+α)2

Squaring both sides.

α2+(α+1)2=(6α)2=(ˆ1+α)2

α2=(6α)2 [ dividing by (α+1)2]

α2=3612α+α2

=α2α2+12α=36

12α=36

α=3612=3


Question 7

The mid-point of the line joining A (2, p) and B (q, 4) is (3, 5). Calculate the numerical values of p and q.

Sol :

Given

(3, 5) is the mid-point of A (2, p) and B (q, 4)

3=2+q2
2+q=6q=62=4

q=4

and 5=p+42p+4=10

p=104=6

p=6

Hence p=6, q=4


Question 8

The ends of a diameter of a circle have the co-ordinates (3, 0) and ( – 5, 6). PQ is another diameter where Q has the coordinates ( – 1, – 2). Find the co-ordinates of P and the radius of the circle.

Sol :

Let AB be the diameter where co-ordinates of

A are (3, 0) and of B are (-5, 6).

Co-ordinates of its origin O will be

(352,0+62) or (22,62) or (-1,3)

Now PQ is another diameter in which co-ordinates of Q are (-1,-2)

Let co-ordinates of P be (x, y) Then co-ordinates of centre O will be

(1+x2,2+y2)

1+x2=1

1+x=2

x=2+1=1

and 2+y2=3

2+y=6

y=6+2=8

Co-ordinates of P will be (-1,8)

Now length of radius OP

=(1+1)2+(83)2

=(0)2+(5)2=0+25

=25=5 units 


Question 9

In what ratio does the point ( – 4, 6) divide the line segment joining the points A( – 6, 10) and B (3, – 8) ?

Sol :

Let the point (-4, 6) divides the line segment joining the points

A (-6, 10) and B (3, -8), in the ratio m : n

4=mx2+nx1m+n=m×3+n(6)m+n

4=3m6nm+n4m4n=3m6n

4n+6n=3m+4m7m=2n

mn=27

Ratio =2 :7


Question 10

Find the ratio in which the point P ( – 3, p) divides the line segment joining the points ( – 5, – 4) and ( – 2, 3). Hence find the value of p.

Sol :

Let P (-3, p) divides AB in the ratio of m1 : m2 coordinates of

A (-5, -4) and B (-2, 3)

3=m1x2+m2x1m1+m2

3=m1(2)+m2(5)m1+m2

3=2m15m2m1+m2

3m13m2=2m15m2

3m1+2m1=5m2+3m2

m1=2m22m2=m1

m1m2=21m1:m2=2:1

Again,

p=m1y2+m2y1m1+m2

=2×3+1×(4)2+1=643=23

Here, p=23


Question 11

In what ratio is the line joining the points (4, 2) and (3, – 5) divided by the x-axis? Also find the co-ordinates of the point of division.

Sol :

Let the point P which is on the x-axis, divides the line segment

joining the points A (4, 2) and B (3, -5) in the ratio of m1 : m2.

and let co-ordinates of P be (x, 0)

0=m1y2+m2y1m1+m2=m1(5)+m2(2)m1+m2

5m1+2m2m1+m2=05m1+2m2=0

5m1=2m25m1=2m2

m1m2=25m1:m2=2:5

Again,

x=m1x2+m2x1m1+m2=2(3)+5(4)2+5=6+207=267

Co-ordinates of P will be (267,0)


Question 12

If the abscissa of a point P is 2, find the ratio in which it divides the line segment joining the points ( – 4 – 3) and (6, 3). Hence, find the co-ordinates of P.

Sol :

Let co-ordinates of A be (-4, 3) and of B (6, 3) and of P be (2, y)

Let the ratio in which the P divides AB be m1:m2

x=m1x2+m2x1m1+m2

2=m1×6+m2×(4)m1+m2

2=6m14m2m1+m2

2m1+2m2=6m14m2

6m12m1=2m2+4m2

4m1=6m2

m1m2=64=32

m1:m2=3:2

y=m1y2+m2y1m1+m2=3×3+2×33+2=9+65=155=3

co-ordinates of P will be (2,3)


Question 13

Determine the ratio in which the line 2x + y – 4 = 0 divide the line segment joining the points A (2, – 2) and B (3, 7). Also find the co-ordinates of the point of division.

Sol :

Points are given A (2, -2), B (3, 7)

and let the line 2x + y – 4 = 0 divides AB in the ratio m1 : m2

at P and let co-ordinates of

x=m1x2+m2x1m1+m2=m1×3+m2×2m1+m2=3m1+2m2m1+m2

and y=m1×7+m2(2)m1+m2=7m12m2m1+m2

P lies on the line 2x+y-4=0, then

2(3m1+2m2)m1+m2+7m12m2m1+m24=0

6m1+4m2+7m12m24m14m2=0

9m12m2=09m1=2m2

m1m2=29 or m1:m2=2:9

x=2×3+2×92+9=6+1811=2411

and y=2×72×92+9=141811=411

Co-ordinates of P will be (2411,411)


Question 14

The point A(2, – 3) is reflected in the v-axis onto the point A’. Then the point A’ is reflected in the line x = 4 onto the:point A”.

(i) Write the coordinates of A’ and A”.

(ii) Find the ratio in which the line segment AA” is divided by the x-axis. Also find the coordinates of the point of division.

Sol :

A’ is the reflection of A(2, -3) in the x-axis

(i) ∴ Co-ordinates of A’ will be (2, 3)

Draw a line x = 4 which is parallel to y-axis

A” is the reflection of A’ (2, 3)

∴Co-ordinates OA” will be (6, 3)


(ii) Join AA” which intersects x-axis at P whose

co-ordinate are (4, 0)

Let P divide AA” in the ratio in m1:m2














y=m1y2+m2y1m1+m2

0=m1×3+m2×(3)m1+m2

3m13m2=03m1=3m2

m1m2=33=11

m1:m2=1:1

Hence P(4, 0) divides AA” in the ratio 1 : 1


Question 15

ABCD is a parallelogram. If the coordinates of A, B and D are (10, – 6), (2, – 6) and (4, – 2) respectively, find the co-ordinates of C.

Sol :

Let the co-ordinates of C be (x, y) and other three vertices

of the given parallelogram are A (10, – 6), B, (2, – 6) and D (4, – 2)

∴ ABCD is a parallelogram

Its diagonals bisect each other.

Let AC and BD intersect each other at O.

∴O is mid-points of BD

∴ Co-ordinates of O will be

(2+42,622) or (62,82) or (3,-4)

Again O is the mid-point of AC then

3=10+x210+x=6

x=610=4

and 4=6+y2

6+y=8y=8+6

y=2

Hence Co-ordinates of C will be (-4,-2)


Question 16

ABCD is a parallelogram whose vertices A and B have co-ordinates (2, – 3) and ( – 1, – 1) respectively. If the diagonals of the parallelogram meet at the point M(1, – 4), find the co-ordinates of C and D. Hence, find the perimeter of the parallelogram. find the perimeter of the parallelogram.

Sol :

ABCD is a || gm , m which co-ordinates of A are (2, -3) and B (-1, -1)

Its diagonals AC and BD bisect each other at M (1, -4)

∴ M is the midpoint of AC and BD

Let co-ordinates of C be(x1,y1) and of D be (x2,y2)

when M is the midpoint of AC then


1=2+x12 and 4=3+y12

2+x1=2

x1=22=0

and 8=3+y1

y1=8+3=5

Co-ordinates of C are (0,-5)

Again M is mid-point of BD, then

1=1+x22,4=1+y22

1+x2=2

x2=2+1=3

and 1+y2=8

y2=8+1=7

co-ordinates of D are (3,-7)

Now, length of AB

=[2(1)]2+[3(1)]2

=(2+1)2+(3+1)2

=(3)2+(2)2=9+4=13

and Length of BC

=[0(1)]2+[5(1)]2

=(1)2+(5+1)2

=1+(4)2=1+16=17

Perimeter of ||g, ABCD=2(AB+BC)

=2(13+17)


Question 17

In the adjoining figure, P (3, 1) is the point on the line segment AB such that AP : PB = 2 : 3. Find the co-ordinates of A and B.











Sol :
A lies on x-axis and
B lies on y-axis
Let co-ordinates of A be (x, 0) and B be (0, y)
and P (3, 1) divides it in the ratio of 2 : 3.

3=m1x2+m2x1m1+m2

=2×0+3×x2+3=0+3x5

3x=15

x=153=5

Again ,

1=m1y2+m2y1m1+m2=2×y+3×02+3

=2y+05=2y5

2y=5y=52

∴Co-ordinates of A will be (5,0) and of B

will be (0,52)


Question 18

Given, O, (0, 0), P(1, 2), S( – 3, 0) P divides OQ in the ratio of 2 : 3 and OPRS is a parallelogram.

Find : (i) the co-ordinates of Q.

(ii)the co-ordinates of R.

(iii) the ratio in which RQ is divided by y-axis.

Sol :
(i) Let co-ordinates of Q be (x’, y’) and of R (x”, y”)
Point P (1, 2) divides OQ in the ratio of 2 : 3

1=m1x2+m2x1m1+m2=2x+3×02+3

2x+05=12x=5

x=52

and 2=m1y2+m2y1m1+m2=2y+3×02+3

2y5=2

2y=10y=5

Co-ordinates of Q will be (52,5)

∵the diagonals of a parallelogram bisect each other

∴In ||gm OPRS, diagonals OR and PS bisect each other at M

∵M is the mid-point of PS

∴Co-ordinates of M will be

=(2+12,0+22) or (22,22) or (-1,1)


(ii) ∵M is the mid-point of OR also

1=0+x2

x=2

and 1=0+y2y=2

Co-ordinates of R will be (-2,2)


(iii) RQ is dividing by y-axis in N

Let the ratio in which N divides RQ in m1:m2

N lies on y-axis

its abscissia (x)=0

0=\frac{m_{2} x^{\prime}+m_{2} x^{\prime \prime}}{m_{1}+m_{2}} $

$\Rightarrow 0=\frac{m_{1} \times \frac{5}{2}+m_{2}(-2)}{\cdot m_{1}+m_{2}}

5m122m2m1+m2=0

52m12m2=0

52m1=2m2

m1m2=2×25=45

m1:m2=4:5


Question 19

If A (5, – 1), B ( – 3, – 2) and C ( – 1, 8) are the vertices of a triangle ABC, find the length of the median through A and the co-ordinates of the centroid of triangle ABC.

Sol :

A (5, -1), B (-3, -2) and C (-1, 8) are the vertices of ∆ABC

D, E and F are the midpoints of sides BC, CA and AB respectively and G is the centroid of the ∆ABC


D is the midpoint of BC
Co-ordinates of D will be

(x1+x22,y1+y22) or (312,2+82)

(42,62) or (-2,3)

G is the centroid

Co-ordinates of G will be

(x1+x2+x33,y1+y2+y33)

or (5313,12+83) or (13,53)


Length of AD=(x2x1)2+(y2y1)2

=(25)2+(3+1)2=(7)2+(4)2

=49+16=65 units

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