# PSEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

PSEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

## PSEB 10th Class Maths Solutions Chapter 7 Coordinate Geometry Ex 7.4

Question 1.

Determine the ratio in which the line it + y – 4 = 0 divides the line segment joining the points A (2, – 2) and B (3, 7).

Solution:

Let line 2x + y – 4 = 0 divides the line segment joining the points A (2,- 2) and B(3, 7) at C (x, y) in the ratio k : 1

Question 2.

Find a relation between x and y if (x, y) ; (1, 2) and (7, 0) are collinear.

Solution:

Let given points are A (x, y); B (1, 2) and C (7, 0).

Here x_{1} = x, x_{2} = 1, x_{3} = 7

y_{1} = y, y_{2} = 2, y_{3} = 0

∵ Three points are collinear

iff 1/2 [x_{1} (y_{2} – y_{3}) + x_{2} (y_{3} – y_{1}) + x_{3} (y_{1} – y_{2})] = 0

or 1/2 x (2 – 0) + 1 (0 – y) + 7 (y – 2)] = 0

or 2x – y + 7y – 14 = 0

or 2x + 6y – 14 = 0

or x + 3y – 7 = 0 is the required relation.

Question 3.

Find the centre of a cirçle passing through the points (6, —6); (3, —7) and (3,3).

Solution:

Let O (x, y) be the required centre of the circle which passes through points P(6, – 6); Q(3, – 7) and R (3, 3).

∴ radii of circle are equal.

∴ OP = OQ = OR

or (OP)^{2} = (OQ)^{2} = (OR)^{2}

Now, (OP)^{2} = (OQ)^{2}

(x – 6)^{2} + (y + 6)^{2} = (x – 3)^{2} + (y + 7)^{2}

or x^{2} + 36 – 12x + y^{2} + 36 + 12y = x^{2} + 9 – 6x + y^{2} + 49 + 14y

or – 12x + 12y + 72 = – 6x + 14y + 58

or – 6x – 2y + 14 = 0

or 3x + y – 7 = 0 ………………(1)

Also, (OQ)^{2} = (OR)^{2}

or (x – 3)^{2} + (y + 7)^{2} = (x – 3)^{2} + (y – 3)^{2}

or (y + 7)^{2} = (y – 3)^{2}

or y^{2} + 49 + 14y = y^{2} + 9 – 6y

or 20y = – 40

y = −40/20 = – 2

Substitute this value of)’ in (1), we get

3x – 2 – 7 = 0

or 3x – 9 = 0

or 3x = 9

or x = 9/3 = 3

∴ Required centre is (3, – 2).

Question 4.

The two opposite vertices of a square are (- 1, 2) and (3, 2). Find the coordinates of other two vertices.

Solution:

Let two opposite vertices of a square ACBD are A (- 1, 2) and B (3, 2) and coordinates of C are (x, y)

∵ Length of each sides of square are equal.

∴ AC = BC

or (AC)^{2} = (BC)^{2}

or (x + 1)^{2} + (y – 2)^{2} = (x – 3)^{2} + (y – 2)^{2}

or (x + 1)^{2} = (x – 3)^{2}

or x^{2} + 1 + 2x = x^{2} + 9 – 6x

or 8x = 8

or x = 8/8 = 1

Now, in rt ∠d ∆ACB,

Using Pythagoras Theorem,

(AC)^{2} + (BC)^{2} = (AB)^{2}

(x + 1)^{2} + (y – 2)^{2} + (x – 3)^{2} + (y – 2)^{2} = (3 + 1)^{2} + (2 – 2)^{2}

or x^{2} + 1 + 2x + y^{2} + 4 – 4y + x^{2} + 9 – 6x + y^{2} + 4 – 4y = 16

or 2x^{2} + 2y^{2} – 4x – 8y + 2 = 0

or x^{2} + y^{2} – 2x – 4y + 1 = 0

Putting the value of x = 1 in (1), we get

(1)^{2} + y^{2} – 2 (1) – 4y + 1 = 0

or y^{2} – 4y = 0

or y (y – 4) = 0

Either y = 0 or y – 4 = 0

Either y = 0 or y = 4

∴ y = 0, 4

∴ Required points are (1. 0) and (1.4).

Question 5.

The Class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Sapling of Gulmohar are planted on the boundary at a distance of 1m from each other. There ¡s a triangular grassy lawn in the plot as shown in the Fig. The students are to sow seeds of flowering plants on the remaining area of the plot.

(i) Taking A as origin, find the coordinates of the vertices of the triangle.

(ii) What will be the coordinates of the vertices of A PQR if C is the origin? Also calculate the areas of the triangles In these cases. What do you observe?

Solution:

Case I:

When taking A as origin then AD is X-axis and AB is Y-axis.

∴ Coordinates of triangular grassy Lawn

PQR are P (4, 6); Q (3, 2) and R(6, 5).

Here x_{1} = 4, x_{2} = 3, x_{3} = 6

y_{1} = 6, y_{2} = 2, y_{3} = 50

Now, area of ∆PQR = 1/2 [x_{1} (y_{2} – y_{3}) + x_{2} (y_{3} – y_{1}) + x_{3} (y_{1} – y_{2})]

= 1/2 [4 (2 – 5) + 3 (5 – 6) + 6 (6 – 2)]

= 1/2 [- 12 – 3 + 24] = 9/2

= 4.5 sq. units.

Case II: When taking C as origin then CB is X – axis and CD is Y – axis.

∴ Coordinates of triangular grassy lawn PQR

are P(12, 2); Q (13,6) and R (10, 3)

Here x_{1} = 12, x_{2} = 13, x_{3} = 10

y_{1} = 2, y_{2} = 6, y_{3} = 3

Now, area of ∆PQR = 1/2 [x_{1} (y_{2} – y_{3}) + x_{2} (y_{3} – y_{1}) + x_{3} (y_{1} – y_{2})]

= 1/2 [12 (6 – 3) + 13 (3 – 2) + 10 (2 – 6)]

= 1/2 [36 + 13 – 40]

= 9/2 = 4.5 sq. units.

From above two cases, it is clear that area of triangular grassy lawn is same.

Question 7.

Let (4, 2), B (6, 5) and C (1, 4) be the vertices of ∆ABC.

(i) The median from A meets BC at D. Find the coordinates of the point D.

(ii) Find the coordinates of the potnt P on AD such that AP : PD = 2 : 1

(iii) Find the coordinates of points Q and R on medians BE and CF respectively such that BQ : QE = 2 : 1 and CR : RF = 2 : 1.

(iv) What do you observe?

[Note : The point which is common to all the three medians ¡s called centroid and this point divides each median in the ratio 2: 1]

(v) if A (x_{1}, y_{1}), B (x_{2}, y_{2}) and C (x_{3}, y_{3}) are the vertices of ∆ABC, find the coordinates of the centroid of the triangle.

Solution:

Given that vertices of ∆ABC are A (4, 2); B (6, 5) and C (1, 4).

(i) AD is the median from the vertex A.

∴ D is the mid point of BC.

(iii) Le BE and CF are the medians of ∆ABC to AC and AB respectively.

∴ E and F are mid points of AC and AB respectively.

(iv) From above discussion, it is clear that coordinates of P, Q and R are same and coincide at a point, is known as centroid of triangle, which divides each median in the ratio 2: 1.

(v) The vertices of given ∆ABC are

A (x_{1}, y_{1}); B (x_{2}, y_{2}) and C (x_{3}, y_{3}).

Question 8.

ABCD is a rectangle formed by the points A (- 1, – 1), B (- 1, 4), C (5, 4) and D (5, – 1). P, Q R and S are the mid points

of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square ? a rectangle? or a rhombus ? Justify your answer.

Solution:

Given: The vertices ot’ given rectangle ABCD are

A(- 1, – 1); B(- 1, 4); C(5, 4) and D (5, – 1).

Form above discussion it is clear that PQ = QR = RS = SP.

Also, PR ≠ QS.

⇒ All sides of quad. PQRS are equal but their diagonals are not equal.

Quad. PQRS is a rhombus.

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