R S Aggarwal and V Aggarwal Solutions for Class 10 Maths CBSE Chapter 6 Co-ordinate Geometry

Class 10^th R. S. Aggarwal Maths Solution

CBSE Class 10 Maths

R. S. Aggarwal Solution

Co-ordinate Geometry Exercise Ex. 6A

Solution 1

(i)The given points are A(9,3) and B(15,11)

(ii)The given points are A(7,4) and B(-5,1)

(iii)The given points are A(-6, -4) and B(9,-12)

(iv)The given points are A(1, -3) and B(4, -6)

(v)The given points are P(a + b, a - b) and Q(a - b, a + b)

(vi)The given points are P(a sin a, a cos a) and Q(a cos a, - a sina)

_{$begin mathsize 12px style left parenthesis straight X subscript 1 equals asin space straight a comma straight y subscript 1 equals straight a space cos space straight a right parenthesis space and space left parenthesis straight x subscript 2 equals acos space straight a comma space straight y subscript 2 equals negative aain space straight a right parenthesis PQ equals square root of left parenthesis straight x subscript 2 minus straight x subscript 1 right parenthesis squared plus left parenthesis straight y subscript 2 minus straight y subscript 1 right parenthesis squared end root space space space space space space equals square root of left parenthesis straight a space cos space straight a space minus space straight a space sin space straight a right parenthesis squared plus left parenthesis negative straight a space cos space straight a right parenthesis to the power of 2 end exponent end root equals square root of straight a squared cos squared straight a plus straight a squared sin squared straight a space minus 2 straight a squared cos space straight a space sin space straight a plus straight a squared cos squared straight a plus straight a squared sin squared straight a plus 2 straight a squared cos space straight a space sin space straight a end root space space space space space equals square root of straight a squared cos squared straight a plus straight a squared sin squared straight a plus straight a squared cos squared end root straight a plus straight a squared sin squared straight a space space space space space equals square root of straight a squared left parenthesis cos squared straight a plus sin squared straight a right parenthesis plus straight a squared left parenthesis cos squared straight a plus sin squared straight a right parenthesis end root space space space space space equals space square root of straight a squared plus straight a squared end root equals square root of 2 straight a squared end root equals square root of 2 straight a space units end style$}

Solution 2

(i)The given point is A(5, -12) and let O(0,0) be the origin

(ii)The given point is B(-5, 5) and let O(0,0) be the origin

(iii)The given point is C(-4, -6) and let O(0,0) be the origin

Solution 3

The given points are A(x, -1) and B(5,3)

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

We know that a point on x-axis is of the form (x, 0).

Let A(x, 0) be the point equidistant from P(-2, 5) and Q(2, -3).

Then,

PA = PB

Hence, the required point is (-2, 0).

Solution 9

Let A(11, -8) be the given point and let P(x,0) be the required point on x - axis

Then,

Hence, the required points are (17,0) and (5,0)

Solution 10

Solution 11

Solution 12

Let A(6, -1) and B(2,3) be the given point and P(x,y) be the required point, we get

Solution 13

Let the required points be P(x,y), then

PA = PB = PC. The points A, B, C are (5,3), (5, -5) and (1, -5) respectively

Hence, the point P is (3, -1)

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18(i)

Solution 18(ii)

Solution 18(iii)

Solution 18(iv)

Solution 19

Solution 20

Solution 21

Solution 22

Solution 23

Let be the given points.

Now,

So, we have

LM = MN = LN

Hence, the triangle LMN formed by the given points is an equilateral triangle.

Solution 24

Let A(-5,6), B(3,0) and C(9,8) be the given points. Then

Solution 25

are the given points

Hence, DABC is equilateral and each of its sides being

Solution 26

(i)The angular points of quadrilateral ABCD are A(3,2), B(0,5), C(-3,2) and D(0,-1)

Thus, all sides of quad. ABCD are equal and diagonals are also equal

Quad. ABCD is a square

(ii)Let A(6,2), B(2,1), C(1,5) and D(5,6) be the angular points of quad. ABCD. Join AC and BD

Thus, ABCD is a quadrilateral in which all sides are equal and the diagonals are equal.

Hence, quad ABCD is a square.

(iii)Let A(0, -2), B(3,1), C(0,4) and D(-3,1) be the angular points of quad. ABCD

Join PR and QSD

Thus, PQRS is a quadrilateral in which all sides are equal and the diagonals are equal

Hence, quad. PQRS is a square

Solution 27

Let A(-3,2), B(-5, -5), C(2, -3) and D(4,4) be the angular point of quad ABCD. Join AC and BD.

Thus, ABCD is a quadrilateral having all sides equal but diagonals are unequal.

Hence, ABCD is a rhombus

Solution 28

Solution 29

Solution 30

Let A(2,1), B(5,2), C(6,4) and D(3,3) are the angular points of a parallelogram ABCD. Then

Diagonal AC Diagonal BD

Thus ABCD is not a rectangle but it is a parallelogram because its opposite sides are equal and diagonals are not equal

Solution 31

Solution 32

(i) Let A(-4, -1), B(-2, -4), C(4, 0) and D(2, 3) are the vertices of quad. ABCD. Then

Thus, ABCD is a quadrilateral whose opposite sides are equal and the diagonals are equal

Hence, quad. ABCD is a rectangle.

(ii)Let A(2, -2), B(14, 10), C(11, 13) and D(-1, 1) be the angular points of quad. ABCD, then

Thus, ABCD is a quadrilateral whose opposite sides are equal and diagonals are equal.

Hence, quad. ABCD is rectangle.

(iii)Let A(0, -4), B(6,2), C(3,5) and D(-3,-1) are the vertices of quad. ABCD. Then

Thus, ABCD is a quadrilateral whose opposite sides are equal and the diagonals are equal

Hence, quad. ABCD is a rectangle

Co-ordinate Geometry Exercise Ex. 6B

Solution 1

(i) The end points of AB are A(-1,7) and B(4, -3)

Let the required point be P(x, y)

By section formula, we have

Hence the required point is P(1, 3)

(ii)The end points of PQ are P(-5, 11) and Q(4, -7)

By section formula, we have

Hence the required point is (2, -3)

Solution 3

Solution 4

Solution 5

Solution 6

Points P, Q, R divide the line segment joining the points A(1,6) and B(5, -2) into four equal parts

Point P divide AB in the ratio 1 : 3 where A(1, 6), B(5, -2)

Therefore, the point P is

Also, R is the midpoint of the line segment joining Q(3, 2) and B(5, -2)

Solution 7

Point P divides the join of A(3, -4) and B(1,2) in the ratio 1 : 2.

Coordinates of P are:

Solution 8

(i)The coordinates of mid - points of the line segment joining A(3, 0) and B(-5, 4) are

(ii)Let M(x, y) be the mid - point of AB, where A is (-11, -8) and B is (8, -2). Then,

Solution 9

The midpoint of line segment joining the points A(6, -5) and B(-2, 11) is

Also, given the midpoint of AB is (2, p)

p = 3

Solution 10

C(1, 2a + 1) is the midpoint of A(2a, 4) and B(-2, 3b)

Solution 11

Let A(-2, 9) and B(6, 3) be the two points of the given diameter AB and let C(a, b) be the center of the circle

Then, clearly C is the midpoint of AB

By the midpoint formula of the co-ordinates,

Hence, the required point C(2, 6)

Solution 12

A, B are the end points of a diameter. Let the coordinates of A be (x, y)

The point B is (1, 4)

The center C(2, -3) is the midpoint of AB

The point A is (3, -10)

Solution 13

Let P divided the join of A(8, 2), B(-6, 9) in the ratio k : 1

By section formula, the coordinates of p are

Hence, the required ratio of which is (3 : 4)

Solution 14

Solution 15

Let P divided the join of line segment A(-4, 3) and B(2, 8) in the ratio k : 1

the point P is

Solution 16

Let P is dividing the given segment joining A(-5, -4) and B(-2, 3) in the ratio r : 1

Coordinates of point P

Solution 17

Let the x- axis cut the join of A(2, -3) and B(5, 6) in the ratio k : 1 at the point P

Then, by the section formula, the coordinates of P are

But P lies on the x axis so, its ordinate must be 0

So the required ratio is 1 : 2

Thus the x - axis divides AB in the ratio 1 : 2

Putting we get the point P as

Thus, P is (3, 0) and k = 1 : 2

Solution 18

Let the y - axis cut the join A(-2, -3) and B(3, 7) at the point P in the ratio k : 1

Then, by section formula, the co-ordinates of P are

But P lies on the y-axis so, its abscissa is 0

So the required ratio is which is 2 : 3

Putting we get the point P as

i.e., P(0, 1)

Hence the point of intersection of AB and the y - axis is P(0, 1) and P divides AB in the ratio 2 : 3

Solution 19

Let the line segment joining A(3, -1) and B(8, 9) is divided by x - y - 2 = 0 in ratio k : 1 at p

Coordinates of P are

Thus the line x - y - 2 = 0 dividesAB in the ratio 2 : 3

Solution 20

Let D, E, F be the midpoint of the side BC, CA and AB respectively in ABC

Then, by the midpoint formula, we have

Hence the lengths of medians AD, BE and CF are given by

Solution 21

Here

Let G(x, y) be the centroid of ABC, then

Hence the centroid of ABC is G(4, 0)

Solution 22

Two vertices of ABC are A(1, -6) and B(-5, 2) let the third vertex be C(a, b)

Then, the co-ordinates of its centroid are

But given that the centroid is G(-2, 1)

Hence, the third vertex C of ABC is (-2, 7)

Solution 23

Two vertices of ABC are B(-3, 1) and C(0, -2) and third vertex be A(a, b)

Then the coordinates of its centroid are

Hence the third vertices A of ABC is A(3, 1)

Solution 24

Let A(3,1), B(0, -2), C(1, 1) and D(4, 4) be the vertices of quadrilateral

Join AC, BD. AC and BD, intersect other at the point O.

We know that the diagonals of a parallelogram bisect each other

Therefore, O is midpoint of AC as well as that of BD

Now midpoint of AC is

And midpoint of BD is

Mid point of AC is the same as midpoint of BD

Hence, A, B, C, D are the vertices of a parallelogram ABCD

Solution 25

Let P(a, -11), Q(5, b), R(2, 15) and S(1, 1) are the vertices of a parallelogram PQRS.

Join the diagonals PR and SQ.

They intersect each other at the point O. We know that the diagonals of a parallelogram bisect each other.

Therefore, O is the midpoint of PR as well as that of SQ

Now, midpoint of PR is

And midpoint of SQ is

Hence the required values are a = 4 and b = 3

Solution 26

Let A(1, -2), B(3, 6) and C(5, 10) are the given vertices of the parallelogram ABCD

Let D(a, b) be its fourth vertex. Join AC and BD.

Let AC and BD intersect at the point O.

We know that the diagonals of a parallelogram bisect each other.

So, O is the midpoint AC as well as that of BD

Midpoint of AC is

Midpoint of BD is

Hence the fourth vertices is D(3, 2)

Solution 27

Solution 28

Solution 29

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Solution 30

Solution 31

Now, O will be the midpoint of AD

Let (t,0) be the co-ordinates of D

Hence,

AO = OD

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Note: The textbook solution is incorrect

Solution 32

Solution 33

Solution 34

Solution 35

A line intersects the y-axis at point P.

∴ Coordinates of P are.(0,y).

Also, the same line intersects the x-axis at point Q.

∴ Coordinates of Q are.(x,0).

Let S be the midpoint of PQ.

∴ Coordinates of S are.(2,-5)

Now,

Therefore, coordinates of P are (0,-10) and coordinates of Q are (4,0).

Solution 36

Let the point R divides PQ in the ratio .

Then, the coordinates of point R are

But, the coordinates of R are given as .

Consider

Hence, the point R divides PQ in the ratio 2:9 and the value of .

Solution 37

Let the coordinates of vertices A, B and C be respectively.

Given, D(3, 4), E(8, 9) and F(6, 7) are midpoints of the sides BC, CA and AB respectively.

Then, we have

From (i), (ii) and (iii), we have

From (i) and (iv),

From (ii) and (iv),

From (iii) and (iv),

Hence, the coordinates of the vertices of triangle ABC are A(11, 12), B(1, 2) and C(5, 6).

Solution 38

Let PQRS be the given paralleogram.

Now, diagonals of a parallleogram bisect each other.

So, we have

Hence, the coordinates of other two vertices of a parallelogram are (1, -12) and (3, -10).

Solution 39

ABCD is a parallelogram.

Then

Coordinates of midpoint of diagonal AC = Coordinates of midpoint of diagonal BD

Solution 40

P and Q are the points of trisection and point P is nearer to A.

⇒ AP:PB = 1:2

Therefore, cordinates of P

Now, point P lies on .

∴ 2(3) - (-2) + = 0

⇒ = -8

Solution 41

Let the required ratio be .

Then, the coordinates of point of division are .

But, it is a point on the y-axis and x-coordinate of every point on y-axis is zero.

Hence, the required ratio is 5:1.

Putting in the coordinates of point of division, we have

Hence, the coordinates of point of division are .

Solution 42

P is the midpoint of points A(3, 4) and B(k, 6).

Now,

Solution 2

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Co-ordinate Geometry Exercise Ex. 6C

Solution 1

(i)Let A(1, 2), B(-2, 3) and C(-3, -4) be the vertices ofthe given ΔABC, then

(ii)The coordinates of vertices of ΔABC are A(-5, 7), B(-4, -5) and C(4, 5)

Here,

$equals 1 half open square brackets negative 5 open parentheses negative 5 minus 5 close parentheses minus 4 left parenthesis 5 minus 7 right parenthesis plus 4 left parenthesis 7 plus 5 right parenthesis close square brackets equals 53 space sq. space units$

(iii)The coordinates of ΔABC are A(3, 8), B(-4, 2) and C(5, -1)

(iv)Let P(10, -6), Q(2, 5) and R(-1, 3) be the vertices of the given ΔPQR. Then,

Solution 2

Solution 3

Solution 4

Solution 5

ABCD is a quadrilateral as follows:

Area of quadrilateral ABCD = Area(ΔABC) + Area(ΔADC)

We have,

Also, we have

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

(i)

(ii)

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