Class 10 R S AGGARWAL AND V AGGARWAL Solutions Maths Chapter 6 - Co-ordinate Geometry

(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)

(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

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

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).

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)

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

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)

6A 

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.

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

are the given points

Hence, DABC is equilateral and each of its sides being

(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

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

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

(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

(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)

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)

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

Coordinates of P are:

(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,

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

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

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)

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)

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)

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

the point P is

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

Coordinates of point P

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

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

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

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

Here

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

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

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)

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)

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

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

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)

Now, O will be the midpoint of AD

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

Hence,

AO = OD

Note: The textbook solution is incorrect

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).

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

Consider

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

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).

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).

ABCD is a parallelogram.

Then

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

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

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 .

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

Now,

(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,

(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,

ABCD is a quadrilateral as follows:

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

We have,

Also, we have

(i)

(ii)

As k > 0

Hence,

k = 3

The given points are A(-3, 12), B(7, 6) and C(x, 9)

Given, points A(-5, 1), B(1, ) and C(4, -2) are collinear.

∴ Area(ΔABC) = 0

Given, points A(, ), B(1, 2) and C(7, 0) are collinear.

∴ Area(ΔABC) = 0

The vertices of ABC are (a, 0), (0, b), C(1, 1)

The points A, B, C are collinear

Area of ABC = 0

ab - a - b = 0 a + b = ab

Dividing by ab

Let  be the given points.

So, we have

It is given that .

∴ Area(ΔABC) ≠ 0.

Hence, the given points will not be collinear. 

Let A(, 3), B(4, 4) and C(3, 5) be the vertices of triangle.

Now, area(ΔABC) = 4

Distance between the points

The points A(4,3) and B(x, 5) lie on the circle with center O(2,3)

OA and OB are radius of the circle.

The point P(x, y) is equidistant from the point A(7, 1) and B(3, 5)

The vertices of ABC are (a, b), (b, c) and (c, a)

Centroid is

But centroid is (0, 0)

a + b + c = 0

The vertices of ABC are A(2, 2), B(-4, -4) and C(5, -8)

Centroid of ABC is given by

Let the point C(4, 5) divides the join of A(2, 3) and B(7, 8) in the ratio k : 1

The point C is

But C is (4, 5)

Thus, C divides AB in the ratio 2 : 3

The points A(2, 3), B(4, k) and C(6, -3) are collinear if area of ABC is zero

But area of ABC = 0,

k = 0

Correct option: (d)

Correct option: (b)

Correct option: (a)