NCERT Solutions Maths Class 10 Exercise 7.4

# NCERT Solutions Maths Class 10 Exercise 7.4

## Maths Class 10 Exercise 7.4

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

Solution: Let the line 2x + y – 4 = 0 divides the line segment joining the points A(2, –2) and B(3, 7) in the ratio k : 1 at point C. Then, the coordinates of C are

But C lies on the line segment 2x + y – 4 = 0, therefore

6k + 4 + 7k – 2 – 4k – 4 = 0

9k – 2 = 0

k = 2/9

Hence, the required ratio is 2 : 9 internally.

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

Solution: We know that the points A(x, y), B(1, 2) and C(7, 0) will be collinear only if the area of triangle ABC = 0.

2x y + 7y – 14 = 0

2x + 6y – 14 = 0

x + 3y – 7 = 0

3. Find the centre of a circle passing through the points (6, –6), (3, –7) and (3, 3).

Solution: Let P(x, y) be the centre of the circle passing through the points A(6, –6), B(3, –7) and C(3, 3). Then AP = BP = CP.

Taking AP = BP

AP2 = BP2

(x 6)2 + (y + 6)2 = (x – 3)2 + (y + 7)2

x2 – 12x + 36 + y2 + 12y + 36 = x2 – 6x + 9 + y2 + 14y + 49

–12x + 6x + 12y – 14y + 72 –58 = 0

–6x – 2y + 14 = 0

3x + y – 7 = 0         ………. (i)

Again, taking BP = CP

BP2 = CP2

(x – 3)2 + (y + 7)2 = (x – 3)2 + (y – 3)2

x2 – 6x + 9 + y2 + 14y + 49 = x2 – 6x + 9 + y2 – 6y + 9

–6x + 6x + 14y + 6y + 58 – 18 = 0

20y + 40 = 0

y = –2

Putting the value of y in equation (i), we get

3x + y – 7 = 0

3x 2 – 7 = 0

3x = 9

x = 3

Hence, the centre of the circle is (3,2).

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

Solution: Let ABCD be a square and B(x, y) be the unknown vertex. A(–1, 2) and B(3, 2) are the known vertices.

Here, AB = BC

AB2 = BC2

(x + 1)2 + (y 2)2  = (x – 3)2 + (y – 2)2

x2 + 2x + 1 + y2 – 4y + 4 = x2 6x + 9 + y2 4y + 4

2x + 1 = 6x + 9

8x = 8

x = 1           ………. (i)

In right-angled ABC, AB2 + BC2 = AC2                [Using Pythagoras theorem]

(x + 1)2 + (y – 2)2 + (x – 3)2 + (y – 2)2 = (3 + 1)2 + (2 – 2)2

2x2 + 2y2 + 2x – 4y – 6x – 4y + 1 + 4 + 9 + 4 = 16

2x2 + 2y2 – 4x – 8y + 2 = 0

x2 + y2 – 2x – 4y + 1 = 0              ………. (ii)

Putting the value of x in equation (ii), we get

1 + y2 – 2 – 4y + 1 = 0

y2 – 4y = 0

y(y –4) = 0

y = 0 or 4

Hence, the required vertices of the square are (1, 0) and (1, 4).

5. The class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Saplings of Gulmohar are planted on the boundary at a distance of 1 m from each other. There is a triangular grassy lawn in the plot as shown in the figure. 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 PQR if C is the origin? Also calculate the area of the triangle in these cases. What do you observe?

Solution: (i) Taking A as the origin, AD and AB are the coordinate axes. Clearly, the coordinates of the points P, Q and R are (4, 6), (3, 2) and (6, 5), respectively.

(ii) Taking C as the origin, CB and CD are the coordinate axes. Clearly, the coordinates of the points P, Q and R are (12, 2), (13, 6) and (10, 3), respectively.

We know that the area of the triangle =

Therefore, area of ∆PQR (First case) = ½ [4(2 – 5) + 3(5 – 6) + 6(6 – 2)]

= ½ [4(–3) + 3(–1) + 6(4)]

= ½ [–12 – 3 + 24

= 9/2 sq. units

And Area of PQR (Second case) = ½ [12(6 – 3) + 13(3 – 2) + 10(2 6)]

= ½ [12(3) + 13(1) + 10(4)]

= ½ [12(3) + 13(1) + 10(4)]

= ½ [36 + 13 40]

= 9/2 sq. units

Hence, the areas are the same in both the cases.

6. The vertices of a ABC are A(4, 6), B(1, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively such that AD/AB = AE/AC = ¼. Calculate the area of the ADE and compare it with the area of ABC.

Solution:  Since, AD/AB = AE/AC = ¼

Therefore, DE BC                  [By Thales theorem]

Therefore, ADE ~ ∆ABC

Therefore,

(AD/AB)2 = (1/4)2 = 1/16        ……….(i)

Now, Area (ABC) = ½ [4(5 – 2) + 1(2 – 6) + 7(6 5)]

= ½ [4(3) + 1(– 4) + 7(1)]

= ½ [12 – 4 + 7] =15/2 sq. units     ………. (ii)

From equations (i) and (ii), we get

Area (ADE) = 1/16 × Area (ABC) = 1/16 × 15/2 = 15/32 sq. units

Therefore, Area (ADE): Area (ABC) = 1 : 16

7. Let A(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 point 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 is called the centroid and this point divides each median in the ratio 2 : 1.]

(v) If A(x1, y1), B(x2, y2) and C(x3, y3) are the vertices of ABC, find the coordinates of the centroid of the triangle.

Solution: It is given that A(4, 2), B(6, 5) and C(1, 4) be the vertices of ABC.

(i) Since AD is the median of ABC from the vertex A.

Therefore, D is the mid-point of BC.

So, its coordinates are

(ii) Since P divides AD in the ratio 2 : 1.

So, the coordinates of point P are (11/3, 11/3)

(iii) Since BE is the median of ABC from the vertex B.

Therefore, E is the mid-point of AC.

So, its coordinates are (5/2, 3).

Since Q divides BE in the ratio 2 : 1.

So, the coordinates of the point Q are  = (11/3, 11/3)

Since CF is the median of ABC from the vertex C.

Therefore, F is the mid-point of AB.

So, its coordinates are = (5, 7/2).

Since R divides CF in the ratio 2 : 1.

So, the coordinates of the point R are  = (11/3, 11/3)

(iv) We observe that the points P, Q and R coincide, i.e., the medians AD, BE and CF are concurrent points. This point is known as the centroid of the triangle. Its coordinates are (11/3, 11/3).

(v) According to the question, D, E and F are the mid-points of BC, CA and AB respectively.

Therefore, the coordinates of D are

Coordinates of a point dividing AD in the ratio 2 : 1 are

The coordinates of E are

Therefore, the coordinates of a point dividing BE in the ratio 2 : 1 are

Similarly, the coordinates of a point dividing CF in the ratio 2 : 1 are

Thus, the point is common to AD, BE and CF and divides them in the ratio 2: 1.

Therefore, the median of a triangle are concurrent and the coordinates of the centroid are  .

8. ABCD is a rectangle formed by joining 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:

Using the distance formula, PQ =

QR =

RS =

SP =

PQ = QR = RS = SP

Now, diagonal PR =

And diagonal SQ =

PR  SQ

Since all the sides are equal but the diagonals are not equal.

Therefore, PQRS is a rhombus.

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