Jharkhand Board JAC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.4 Textbook Exercise Questions and Answers.

## JAC Board Class 10 Maths Solutions Chapter 7 Coordinate Geometry Exercise 7.4

Question 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).

Answer:

Let the line 2x + y – 4 = 0 divide the line segment joining the point A (2, – 2) and B (3, 7) at point M (a, b) in the ratio k : 1.

Then, the coordinates of point M are

(\(\frac{3 k+2}{k+1}, \frac{7 k-2}{k+1}\)) = (a, b)

∴ a = \(\frac{3k+2}{k+1}\)

and b = \(\frac{7k-2}{k+1}\) …… (1)

Now, the point M also lies on line 2x + y – 4 = 0.

Hence, (a, b) must satisfy

2x + y – 4 = 0

∴ 2a + b – 4 = 0

∴ 2(\(\frac{3k+2}{k+1}\)) + (\(\frac{7k-2}{k+1}\)) – 4 = 0 [From (1)]

∴ 6k + 4 + 7k – 2 – 4k – 4 = 0

∴ 9k – 2 = 0

∴ k = \(\frac{2}{9}\)

Hence, the ratio k : 1 = \(\frac{2}{9}\) : 1 = 2 : 9.

Thus, the line 2x + y – 4 = 0 divides the line segment joining the points A (2, – 2) and B (3, 7) in the ratio 2 : 9.

Question 2.

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

Answer:

Points A (x, y), B(1, 2) and C(7,0) are collinear.

∴ Area of ΔABC = 0

∴ \(\frac{1}{2}\)[x(2 – 0) + 1 (0 – y) + 7 (y – 2)] = 0

∴ 2x – y + 7y – 14 = 0

∴ 2x + 6y – 14 = 0

∴ x + 3y – 7 = 0

Thus, x + 3y – 7 = 0 is the required relation between x and y.

Question 3.

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

Answer:

Let A (6, 6), B (3, 7) and C(3, 3) be the given points and P(x, y) be the centre of the circle passing through these points. Then, P is equidistant from all the points A, B and C.

∴ PA = PB = PC

∴ PA² = PB² = PC²

Now, PA² = PB² gives

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

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

∴ – 6x – 2y = – 14

∴ 3x + y = 7 …………..(1)

Again, PA² = PC² gives

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

∴ x² – 12x + 36 + y² + 12y + 36 = x² – 6x + 9 + y² – 6y + 9

∴ – 6x + 18y = – 54

∴ x – 3y = 9 …………..(2)

Multiplying equation (1) by 3, we get

9x + 3y = 21

Adding equations (2) and (3), we get 10x = 30

10x = 3

∴ x = 3

Substituting x = 3 in 3x + y = 7, we get 9 + y = 7. i.e., y = – 2

∴ The coordinates of P are (3, – 2).

Thus, the centre of the circle passing through the given points is (3, – 2).

Question 4.

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

Answer:

Let ABCD be the given square in which A(-1, 2) and C(3, 2) are given vertices and B and D are to be found.

Let the coordinates of B be (x, y).

In ΔABC, ∠B = 90° and AB = BC.

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

∴ 8x = 8

∴ x = 1

In ΔABC,

∠B = 90°

Hence, by Pythagoras theorem,

AB² + BC² = AC²

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

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

∴ 2x² – 4x + 2y² – 8y + 18 – 16 = 0

∴ 2(1)² – 4(1) + 2y² – 8y + 2 = 0 (Substituting x = 1)

∴ 2 – 4 + 2y² – 8y + 2 = 0

∴ 2y² – 8y = 0

∴ 2y (y – 4) = 0

∴ 2y = 0 or y – 4 = 0

∴ y = 0 or y = 4

These two values of y gives us the y-coordinates of B and D while the single value of x suggests that the x-coordinate of B and D is same. Thus, the coordinates of the other two vertices are (1, 0) and (1, 4) respectively.

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 1 m from each other. There is a triangular grassy lawn in the plot as shown in the given figure. The students are to sow seeds of flowering plants on the remaining area of the plot.

1. Taking A as origin, find the coordinates of the vertices of the triangle.

2. What will be the coordinates of the vertices of ΔPQR if C is the origin? Also calculate the areas of the triangles in these cases. What do you observe?

Answer:

1. Taking A as the origin and AB and AD as the y-axis and the x-axis respectively, we get the coordinates of points as A (0, 0), B(0, 8), D(16, 0) and C(16, 8).

Then, the vertices of ΔPQR are P (4, 6), 9 (3, 2) and R (6, 5).

Area of ΔPQR

= \(\frac{1}{2}\)[4(2 – 5) + 3(5 – 6) + 6(6 – 2)]

= \(\frac{1}{2}\)[- 12 – 3 + 24]

= \(\frac{9}{2}\) sq units

2. Taking C as the origin and CB and CD as positive x-axis and y-axis respectively, we get C(0, 0), B(16, 0), D (0, 8) and A (16, 8) as the vertices of quadrilateral ABCD. Then, the vertices of ΔPQR are P (12, 2), Q (13, 6) and R(10, 3).

Area of ΔPQR

= \(\frac{1}{2}\)[12(6 – 3) + 13 (3 – 2) + 10 (2 – 6)]

= \(\frac{1}{2}\)[36 + 13 – 40]

= \(\frac{9}{2}\)sq units

We observe that the area of ΔPQR is same in both the cases.

Note: In second case, if we take CB and CD as negative x-axis and y-axis, then the vertices of ΔPQR will be P(-12, -2), Q(-13,-6) and R(-10,-3). Still the area of ΔPQR would not change.

Question 6.

The vertices of a ΔABC 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 \(\frac{AD}{AB}=\frac{AE}{AC}=\frac{1}{4}\). Calculate the area of the ΔADE and compare it with the area of ΔABC.

Answer:

Similarly, \(\frac{AE}{EC}=\frac{1}{3}\)

Now, D divides the join of A (4, 6) and B(1, 5) in the ratio 1:3.

Then the coordinates of D are

So, area of ΔADE : area of ΔABC = 1 : 16

Thus, the area of ΔADE is \(\frac{15}{32}\) sq units and the ratio of areas of ΔADE and ΔABC is 1 : 16.

Question 7.

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

1. The median from A meets BC at D. Find the coordinates of the point D.

2. Find the coordinates of the point P on AD such that AP : PD = 2 : 1

3. 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.

4. 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.]

5. If A(x_{1}, y_{1}), B(x_{2}, y_{2}) and C(x_{3}, y_{3}) and the vertices of ΔABC, find the coordinates of the centroid of the triangle.

Answer:

1. AD is a median. Hence, D is the midpoint of BC.

Then, the coordinates of D, by midpoint formula, are (\(\frac{6+1}{2}, \frac{5+4}{2}\))

= (\(\frac{7}{2}\), \(\frac{9}{2}\)).

2. Point P divides AD in the ratio 2 : 1.

∴ Coordinates of P by section formula, are (\(\frac{2\left(\frac{7}{2}\right)+1(4)}{2+1}, \frac{2\left(\frac{9}{2}\right)+2}{2+1}\)) = (\(\frac{11}{3}\), \(\frac{11}{3}\))

3. BE is a median. Hence, E is the midpoint of AC.

Then, the coordinates of E, by midpoint formula, are

(\(\frac{4+1}{2}, \frac{2+4}{2}\)) = (\(\frac{5}{2}\), 3).

Now, Q divides BE in the ratio 2 : 1.

∴ Coordinates of Q, by section formula, are (\(\frac{2\left(\frac{5}{2}\right)+1(6)}{2+1}, \frac{2(3)+1(5)}{2+1}\)) = (\(\frac{11}{3}\), \(\frac{11}{3}\))

CF is a median, Hence, F is the midpoint of AB.

Then, the coordinates of F, by midpoint formula are (\(\frac{4+6}{2}, \frac{2+5}{2}\)) = (5, \(\frac{7}{2}\)).

Point R divide CF in the ratio 2 : 1.

∴ Coordinates of R, by section formula, are (\(\frac{2(5)+1(1)}{2+1}, \frac{2\left(\frac{7}{2}\right)+1(4)}{2+1}\)) = (\(\frac{11}{3}\), \(\frac{11}{3}\))

4. Here, we observe that points P(\(\frac{11}{3}\), \(\frac{11}{3}\)), Q(\(\frac{11}{3}\), \(\frac{11}{3}\)) and R(\(\frac{11}{3}\), \(\frac{11}{3}\)) are the same points. i.e., P = Q = R.

This means that all the three medians are concurrent at point P (or Q or R).

This point is called the centroid of the triangle which is usually denoted by letter G. Centroid of a triangle divides each of its medians in the ratio 2 : 1 from the vertex side.

5. Now, the coordinates of the vertices of ΔABC are A(x_{1}, y_{1}), B(x_{2}, y_{2}) and C(x_{3}, y_{3}).

AD, BE and CF are medians.

Hence, D, E and F are the midpoints of BC, AC and AB respectively.

Note: Now, with this formula, we can verify the answers received in part (2) and (3), as (\(\frac{4+6+1}{3}, \frac{2+5+4}{3}\)) = (\(\frac{11}{3}\), \(\frac{11}{3}\))

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

Answer:

In rectangle ABCD, P, Q, R and S are the midpoints of AB, BC, CD and DA respectively. Then, the coordinates of P, Q, R and S are obtained as

So, in quadrilateral PQRS,

PQ² = QR² = RS² SP2 but PR² ≠ QS², i.e..

PQ = QR = RS = SP but PR ≠ QS

So, in quadrilateral PQRS, all the sides are equal but diagonals are not equal.

Hence, PQRS is a rhombus.