Bihar Board Class 10th Maths Chapter 7 Coordinate Geometry Solution

 

BSEB Bihar Board Class 10th Maths Solutions Chapter 7 Coordinate Geometry Ex 7.1

Question 1.
Find the distance between the following pairs of points :
(i) (2, 3), (4, 1)
(ii) (- 5, 7), (- 1, 3)
(iii) (a, b), (- a, – b)
Solution:
(i) Let P(2, 3) and Q(4, 1) be the given points.
Here, x1 = 2, y1 = 3 and x2 = 4, y2 = 1.
∴ PQ = (x2−x1)2+(y2−y1)2−−−−−−−−−−−−−−−−−−√
So, PQ = (4−2)2+(1−3)2−−−−−−−−−−−−−−−√ = (2)2+(−2)2−−−−−−−−−−√
or PQ = 4+4−−−−√ = 8–√ = 22–√

(ii) Let P(- 5, 7) and Q(- 1, 3) be the given points.
Here, x1 = -5, y1 = 7 and x2 = -1, y2 = 3.
∴ PQ = (x2−x1)2+(y2−y1)2−−−−−−−−−−−−−−−−−−√
So, PQ = (−1+5)2+(3−7)2−−−−−−−−−−−−−−−−√ = (4)2+(−4)2−−−−−−−−−−√
or PQ = 16+16−−−−−−√ = 32−−√ = 16×2−−−−−√ = 42–√

(iii) Let P(a, b) and Q(- a, – b) be the given points.
Here, x1 = a, y1 = b and x2 = a, y2 = – b.
∴ PQ = (x2−x1)2+(y2−y1)2−−−−−−−−−−−−−−−−−−√
So, PQ = (−a−a)2+(−b−b)2−−−−−−−−−−−−−−−−−−√
= (−2a)2+(−2b)2−−−−−−−−−−−−−√
or PQ = 4a2+4b2−−−−−−−−√
= 2a2+b2−−−−−−√

Question 2.
Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2?
Solution:
Let P(0, 0) and Q(36, 15) be the given points.
Here, x1 = 0, y1 = 0 and x2 = 36, y2 = 15.
∴ PQ = (x2−x1)2+(y2−y1)2−−−−−−−−−−−−−−−−−−√
So, PQ = (36−0)2+(15−0)2−−−−−−−−−−−−−−−−−√ = 1296+225−−−−−−−−−√
= 1521−−−−√ = 39
In fact, the positions of towns A and B are given by (0, 0) and (36, 15) respectively and so the distance between them = 39 km as calculated above.

Question 3.
Determine if the points (1, 5), (2, 3) and (- 2, – 11) are collinear.
Solution:
Let A(1, 5), B(2, 3) and C(- 2, – 11) be the given points. Then, we have :

Clearly, BC ≠ AB + AC, AB ≠ BC + AC and AC ≠ BC + AB.
Hence, A, B and C are not collinear.

Question 4.
Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle.
Solution:
Let A(5, – 2), B(6, 4) and C(7, – 2) are the given points. Then,
AB = (6−5)2+(4+2)2−−−−−−−−−−−−−−−√
= 1+36−−−−−√
= 37−−√
BC = (7−6)2+(−2−4)2−−−−−−−−−−−−−−−−√
= 1+36−−−−−√
= 37−−√
Clearly, AB = BC
∴ ∆ ABC is an isosceles triangle.

Question 5.
In a classroom, four friends are Seated at the points A, B, C and D as shown in figure. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square ?” Chameli disagrees. Using distance formula, find which of them is correct, and why?
Solution:
Clearly from the figure, the coordinates of points A, B, C and D are (3, 4), (6, 7), (9, 4) and (6, 1).
By using distance formula, we get

Thus, the four sides are equal and the diagonals are also equal. Therefore, ABCD is a square.
Hence, Champa is correct.

Question 6.
Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer :
(i) (- 1, – 2), (1, 0), (- 1, 2), (- 3, 0)
(ii) (-3, 5), (3, 1), (0, 3), (- 1,-4)
(iii) (4, 5), (7, 6), (4, 3), (1, 2)
Solution:
(i) Let A(- 1, – 2), B(1, 0), C(- 1, 2) and D(- 3, 0) be the given points. Then,

Clearly, four sides AB, BC, CD and DA are equal.
Also, diagonals AC and BD are equal.
∴ The quadrilateral ABCD is a square.

(ii) Let A(- 3, 5), B(3, 1), C(0. 3) and D(- 1, – 4) be the given points. Plot these points as shown.

Clearly, the points A, C and B are collinear.
So, no quadrilateral is formed by these points.

Aliter :
AB = (3+3)2+(1−5)2−−−−−−−−−−−−−−−√
= 36+16−−−−−−√
= 52−−√
= 213−−√

BC = (3−0)2+(1−3)2−−−−−−−−−−−−−−−√
and AC = (0+3)2+(3−5)2−−−−−−−−−−−−−−−√ = 13−−√
So, AC = AB + BC.
Hence, there, will not be any quadrilateral.

(iii) Let A(4, 5), B(7, 6), C(4, 3) and D(1, 2) be the .given points. Then,

Clearly, AB = CD, BC = DA and AC = BD.
∴ The quadrilateral ABCD is a parallelogram.

Question 7.
Find the point on the x-axis which is equidistant from (2, – 5) and (- 2, 9).
Solution:
Since the point on x-axis have its ordinate – 0, so P(x, 0) is any point on the x-axis.
Since P(x, 0) is equidistant from A(2, – 5) and P (- 2, 9), therefore PA = PB gives. PA² = PB²
or (x – 2)² + (0 + 5)² = (x + 2)² + (0 – 9)²
or x² – 4x + 4 + 25 = x² + 4x + 4 + 81
or – 4x – 4x = 81 – 25
– 8x = 56
or x = 56−8 = – 7
∴ The point equidistant from given points on the x-axis is (- 7, 0).

Question 8.
Find the values of y for which the distance between the points P(2, – 3) and Q(10, y) is 10 units.
Solution:
P(2, – 3) and Q(10, y) are given points such that PQ = 10 units.

Thus, the possible value of y is – 9 or 3.

Question 9.
If Q(0, 1) is equidistant from P(5, – 3) and R(x, 6), find the values of x. Also, find the distances QR and PR.
Solution:
Since the point Q(0, 1) is equidistant from P(5, – 3) and R(x, 6), therefore
QP = QR gives QP² – QR²
or (5 – 0)² + (- 3 – 1)² = (x – 0)² + (6 – 1)²
or 25 + 16 = x² + 25
or x² = 16 So, x = ± 4
Thus, R is (4, 6) or (- 4, 6).
Now, QR = Distance between Q(0, 1) and R(4, 6)
(4−0)2+(6−1)2−−−−−−−−−−−−−−−√ = 16+25−−−−−−√ = 41−−√
Also, QR = Distance between Q(0, 1) and R(- 4, 6)
(−4−0)2+(6−1)2−−−−−−−−−−−−−−−−√ = 16+25−−−−−−√ = 41−−√
and PR = Distance between P(5, – 3) and R(4, 6)
(4−5)2+(6+3)2−−−−−−−−−−−−−−−√ = 1+81−−−−−√ = 82−−√
and PR = Distance between P(5, – 3) and R(-4, 6)
(−4−5)2+(6+3)2−−−−−−−−−−−−−−−−√ = 81+81−−−−−−√ = 92–√

Question 10.
Find a relation between x and y such that the point Or, y) is equidistant from the points (3, 6) and (-3,4).
Solution:
Let the point P(x, y) be equidistant from the points A(3, 6) and B(- 3, 4).
i.e., PA = PB gives PA² = PB²
or (x – 3)² + (y – 6)² = (x + 3)² + (y – 4)²
or x²- 6x + 9 + y² – 12y + 36 = x²+ 6x + 9 + y² – 8y + 16
or 6x – 6x – 12y + 8y + 36 – 16 0
or – 12x – 4y + 20 = 0
or 3x + y – 5 = 0,
which is the required relation.

BSEB Bihar Board Class 10th Maths Solutions Chapter 7 Coordinate Geometry Ex 7.2

Question 1.
Find the coordinates of the point which divides the join of (- 1, 7) and (4, – 3) in the ratio 2:3.
Solution:
Let P(x, y) be the required point.

So, the coordinates of P are (1, 3).

Question 2.
Find the coordinates of the points of trisection of the lire segment joining (4, – 1) and (- 2, – 3).
Solution:
Let P and Q be the points of trisection of the line segment joining A(4, – 1) and B(- 2, – 3). Then, AP = PQ = QB = k (say).
∴ PB = PQ + QB = 2k
and AQ = AP + PQ = 2k
So, AP : PB = k : 2k = 1 : 2 and AQ : QB = 2k : k = 2 : 1

So, P divides AB internally in the ratio 1 : 2, while Q divides AB internally in the ratio 2 : 1. Thus, the coordinates of P are

Hence, the two points of trisection are
(2, \(\frac { -5 }{ 3 }\)) and (0, \(\frac { -7 }{ 3 }\)).

Question 3.
To conduct Sports Day activities in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1 m from each other along AD, as shown in figure.

Niharika runs \(\frac { 1 }{ 4 }\) th the distance AD on the 2nd line and posts a green flag. Preet runs \(\frac { 1 }{ 5 }\) th the distance AD on the eighth line and posts a red flag. What is the distance between both the flags ? If Rashmi has to post a blue flag exactly halfway between the line (segment) joining the two flags, where should she post her flag ?
Solution:
Clearly from the figure, the position of green flag posted by Niharika is given by P(2, \(\frac { 1 }{ 4 }\) x 100) i.e., P(2, 25) and that of red, flag posted by Preet is given by Q(8, \(\frac { 1 }{ 5 }\) x 100), i.e., Q(8, 20)
Now, PQ = \(\sqrt{(8-2)^{2}+(20-25)^{2}}\) = \(\sqrt{(6)^{2}+(-5)^{2}}\)
= \(\sqrt{36+25}\)
= \(\sqrt{61}\)
∴ The distance between the flags = \(\sqrt{61}\) metres
Let M be the position of the blue flag posted by Rashmi in the halfway of line segment PQ.

∴ M is the given by \(\left(\frac{2+8}{2}, \frac{25+20}{2}\right)\)
or \(\left(\frac{10}{2}, \frac{45}{2}\right)\), i.e., (5, 22.5)
Thus, the blue flag should be on the fifth line at a distance of 22.5 m above it.

Question 4.
Find the ratio in which the line segment joining the points (- 3, 10) and (6, – 8) is divided by (- 1, 6).
Solution:
Let the point P(- 1, 6) divide the line joining A(- 3, 10) and B(6, – 8) in- the ratio k : 1. Then, the coordinates of P are \(\left(\frac{6 k-3}{k+1}, \frac{-8 k+10}{k+1}\right)\)
But, the coordinates of P are given as (- 1, 6).

Hence, the point P divides AB in the ratio 2 : 7.

Question 5.
Find the ratio in which the line segment joining A(1, – 5) and B(- 4, 5) is divided by the x-axis. Also, find the coordinates of the point of division.
Solution:
Let the required ratio be k : 1. Then, the coordinates of the point P of division are \(\left(\frac{-4 k+1}{k+1}, \frac{5 k-5}{k+1}\right)\)

But it is a point on x-axis on which y-coordinate of every point is 0.
∴ \(\frac{5 k-5}{k+1}\) = 0
or 5k – 5 = 0
5k = 5
k = 1
Thus, the required ratio is 1 : 1 and the point of division P is given by
\(\left(\frac{-4 \times 1+1}{1+1}, 0\right)\), i.e., \(\left(\frac{-4+1}{2}, 0\right)\) i.e., \(\left(\frac{-3}{2}, 0\right)\)

Question 6.
If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.
Solution:
Let A(1, 2), B(4, y), C(x, 6) and D(3, 5) be the vertices of a parallelogram ABCD.
Since the diagonals of a parallelogram bisect each other, therefore

\(\frac{x+1}{2}\) = \(\frac{3+4}{2}\)
or x + 1 = 7 or x = 6
and \(\frac{5+y}{2}\) = \(\frac{6+2}{2}\)
or 5 + y = 8
or y = 3
Hence, x = 6 and y = 3.

Question 7.
Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, – 3) and B is (1, 4).
Solution:
Let AB be a diameter of the circle having its centre at C(2, – 3) such that the coordinates of end B are (1, 4).

Let the coordinates of A be (x, y).
Since C is the mid-point of AB, therefore
the coordinates of C are \(\left(\frac{x+1}{2}, \frac{y+4}{2}\right)\)
But, the coordinates of C are given as (2, – 3).
∴ \(\frac{x+1}{2}\) = 2 or x + 1 = 4 or x = 3
and \(\frac{y+4}{2}\) = – 3 or y + 4 = – 6 or y = – 10
Hence, the coordinates of A are (3, – 10).

Question 8.
If A and B are (- 2, – 2) and (2, – 4) respectively, find the coordinates of P such that AP = lies on the line segment AB.
Solution:
We have : AP = \(\frac{3}{7}\)AB

Let P(x, y) be the point which divides the join of A(- 2, – 2) and B(2, – 4) in the ratio 3:4.

Hence, the coordinates of the point P are \(.

Question 9.
Find the coordinates of the points which divide the line segment joining A(- 2, 2) and B(2, 8) into four equal parts.
Solution:
Let P1 P2 and P3 be the points that divide the line segment joining A(- 2, 2) and B(2, 8) into four equal parts.

Since P1 divides the line segment into two equal parts, therefore
coordinates of P2 (i.e., mid-point) are [latex]\left(\frac{-2+2}{2}, \frac{2+8}{2}\right)\), i.e., (0, 5).
Now, P1 divides the line segment AP2 into two equal parts.
Coordinates of P1 (i.e., mid-point) are \(\left(\frac{-2+0}{2}, \frac{2+5}{2}\right)\), i,e., \(\left(-1, \frac{7}{2}\right)\),
Again, P3 is the mid-point of line segment P2B.
∴ Coordinates or P3 are \(\left(\frac{0+2}{2}, \frac{5+8}{2}\right)\), i.e., (1, \(\frac{13}{2}\)).

Question 10.
Find the area of a rhombus if its vertices are (3, 0), (4, 5), (- 1, 4) and (- 2, – 1) taken in order.
Solution:
Let A(3, 0), B(4, 5), C(- 1, 4) and D(- 2, – 1) be the vertices of the rhombus ABCD.

BSEB Bihar Board Class 10th Maths Solutions Chapter 7 Coordinate Geometry Ex 7.3

Question 1.
Find the area of the triangle whose vertices are :
(i) (2, 3), (- 1, 0), (2, – 4)
(ii) (- 5, – 1), (3, – 5), (5, 2)
Solution:
(i) Let A = (x1, y1) = (2, 3), B = (x2, y2) = (- 1, 0) and C = (x3, y3) = (2, – 4).

(ii) Let A = (x1, y1) = (- 5, – 1), B = (x2, y2) = (3, – 5) and C = (x3, y3) = (5, 2).

Question 2.
In each of the following, find the value of ‘k\ for which the points are collinear :
(i) (7, – 2), (5, 1), (3, k)
(ii) (8, 1), (k, – 4), (2, – 5)
Solution:
(i) Let the given points be A = (x1, y1) = (7, – 2), B = (x2, y2) = (5, 1) and C = (x3, y3) = (3, k). These points lie on a line if
Area (∆ ABC) = O
i.e., x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) = 0
or 7(1 – k) + 5(k + 2) + 3(- 2 – 1) = 0
or 7 – 7k + 5k + 10 – 9 = 0
or 8 – 2k = 0
2k = 8 i.e., k = 4
Hence, the given points are collinear for k = 4

(ii) Let the given points be A = (x1, y1) = (8, 1), B = (x2, y2) = (k, – 4) and C = (x3, y3) = (2, – 5).
If the given points are collinear, then
x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) = 0= 0
So, 8(- 4 + 5) + k(- 5 – 1) + 2(1 + 4) = 0
or 8 – 6k + 10 = 0
or – 6k = – 18
i.e., k – 3
Hence, the given points are collinear for k = 3.

Question 3.
Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, – 1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.
Solution:
Let A = (x1, y1) = (0, – 1), B = (x2, y2) = (2, 1) and C = (x3, y3) = (0, 3) be the vertices of ∆ ABC.

Let P \(\left(\frac{0+2}{2}, \frac{3+1}{2}\right)\) i.e., (1, 2),
Q \(\left(\frac{2+0}{2}, \frac{1-1}{2}\right)\) i.e., (1, 0) and ,
R \(\left(\frac{0+0}{2}, \frac{3-1}{2}\right)\) i.e. (0, 1) be the
vertices of A PQR formed by joining the mid-points of the sides of A ABC.
Now,

Ratio of the area (∆ PQR) to the area (∆ ABC) = 1 : 4.

Question 4.
Find the area of the quadrilateral whose vertices, taken in order, are (- 4, – 2), (- 3, – 5), (3, – 2) and (2, 3).
Solution:
Let A(- 4, – 2), B(- 3, – 5), C(3, – 2) and D(2, 3) be the vertices of the quadrilateral ABCD.
Area of quadrilateral ABCD = Area of ∆ ABC + Area of ∆ ACD
= \(\frac { 1 }{ 2 }\)[- 4(- 5 + 2) – 3(- 2 + 2) + 3(- 2 + 5)] + \(\frac { 1 }{ 2 }\)[- 4(- 2 – 3) + 3(3 + 2) + 2(- 2 + 2)]
= \(\frac { 1 }{ 2 }\)(12-0 + 9) + \(\frac { 1 }{ 2 }\)(20 + 15 + 0)
= \(\frac { 1 }{ 2 }\)(21 + 35) sq. units = \(\frac { 1 }{ 2 }\) x 56 sq.units = 28 sq. units

Question 5.
You have studied in class IX (Chapter 9, Example 3) that a median of a triangle divides it into two triangles of equal areas. Verify this result for ∆ ABC whose vertices are A(4, – 6), B(3, – 2) and C(5, 2).
Solution:

Since AD is the median of A ABC, therefore D is the mid-point of BC. Coordinates of D are \(\left(\frac{3+5}{2}, \frac{-2+2}{2}\right)\), i.e., (4, 0)
Area of ∆ ADC = \(\frac { 1 }{ 2 }\) [4(0 – 2) + 4(2 + 6) + 5(- 6 – 0)]
= \(\frac { 1 }{ 2 }\)(- 8 + 32 – 30) = \(\frac { 1 }{ 2 }\) x (- 6) = – 3
= 3 sq. units (numerically)
Area of ∆ ABD = \(\frac { 1 }{ 2 }\) [4(- 2 – 0) + 3(0 + 6) + 4(- 6 + 2)]
= \(\frac { 1 }{ 2 }\)(- 8 + 18 – 16)
= \(\frac { 1 }{ 2 }\)(- 6)
= – 3
= 3 sq. units (numerically)
Clearly, area (∆ ADC) = area (∆ ABD).
Hence, the median of the triangle divides it into two triangles of equal areas.

BSEB Bihar Board Class 10th Maths Solutions Chapter 7 Coordinate Geometry Ex 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).
Solution:
Suppose the line 2x + y – 4 = 0 divides the line segment joining A(2, – 2) and B(3, 7) in the ratio k : 1 at point C. Then, the coordinates of C are \(\left(\frac{3 k+2}{k+1}, \frac{7 k-2}{k+1}\right)\)

But C lies on 2x + y – 4 = 0. Therefore,
2\(\left(\frac{3 k+2}{k+1}\right)\) + \(\left(\frac{7 k-2}{k+1}\right)\) – 4 = 0
or 6k + 4 + 7k – 2 – 4k – 4 = 0
or 9k – 2 = 0
or 9k = 2 i.e., \(\frac { 2 }{ 9 }\)
So, the required ratio is 2 : 9 internally.

Question 2.
Find a relation between x and y, if the points (x, y), (1, 2) and (7, 0) are collinear.
Solution:
The points A(x, y), B(1, 2) and C(7, 0) will be collinear, if
x(2 – 0) + 1(0 – y) + 7(y – 2) = 0
i.e., 2x – y + 7y – 14 = 0
or 2x + 6y – 14 = 0
or x + 3y – 7 = 0
which is the required relation between x and y.

Question 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.
Now, AP = BP
So, AP² = BP²
So, (x – 6)² + (y + 6)² = (x – 3)² + (y + 7)²
or x² – 12x + 36 + y² + 12y + 36 = x² – 6x + 9 + y² + 14y + 49 or – 12x + 6x + 12y – 14y + 72 – 58 = 0
or – 6x – 2y + 14 = 0
or 3x +y – 7 = 0 … (1)
Also, BP = CP
So, BP² = CP²
So, (x – 3)² + (y + 7)² = (x – 3)² + (y – 3)²
or x² – 6x + 9 + y² + 14y + 49 = x² – 6x + 9 + y² – 6y + 9
or – 6x + 6x + 14y + 6y + 58 – 18 = 0
or 20y + 40 = 0
or y = \(\frac { -40 }{ 20 }\) = – 2 …. (2)
Putting y = – 2 in (1), we get
3x – 2 – 7 = 0
or 3x = 9
or x = 3
Thus, the centre of the circle 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.
Solution:
Let ABCD be a square and let A(- 1, 2) and C(3, 2) be the given opposite vertices.
Let B(x, y) be the unknown vertex.

Then, AB = BC gives AB² = BC²
So, (x + 1)² + (y – 2)² = (x – 3)² + (y – 2)²
or 2x + 1 = – 6x + 9
or 8x = 8
or x = 1 … (1)
In A ABC, we have : AB² + BC² = AC2. A(“
So, (x + 1)² + (y – 2)² + (x – 3)² + (y – 2)² = (3 + 1)² + (2 – 2)²
or 2x² + 2y² + 2x – 4y – 6x – 4y + 1 + 4 + 9 + 4 = 16
or 2x² + 2y² – 4x – 8y + 2 = 0
or x² + y² – 2x – 4y + 1 = 0 … (2)
Substituting the value of x from (1) in (2), we get
1 + y² – 2 – 4y + 1 = 0
or y² – 4y = 0
or y(y – 4) = 0
i. e., y – 0 or y = 4
Hence, the required vertices of the square 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. 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 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 points P, Q and R are given by (12, 2), (13, 6) and (10, 3) respectively.
We know that the area of the triangle whose vertices are (x1, y1), (x2, y2) and (x3, y3) is given by
\(\frac { 1 }{ 2 }\)[x1(y2 -y3) + x2(y3 -y1) + x3(y1 – y2)].
∴ Area of ∆ PQR in the 1st case
= \(\frac { 1 }{ 2 }\)[4(2 – 5) + 3(5 – 6) + 6(6 – 2)]
= \(\frac { 1 }{ 2 }\)(4 x (-3) + 3 x (- 1) + 6 x 4)
= \(\frac { 1 }{ 2 }\)(- 12 – 3 + 24) = \(\frac { 9 }{ 2 }\)sq. units
and Area of ∆ PQR in the 2nd case
= \(\frac { 1 }{ 2 }\)[12(6 – 3) + 13(3 – 2) + 10(2 – 6)]
= \(\frac { 1 }{ 2 }\)(12 x 3 + 13 x 1 + 10 x (-4))
= \(\frac { 1 }{ 2 }\)(36 + 13 – 40) = \(\frac { 9 }{ 2 }\)sq. units
Thus, we observe that the areas are the same in both the cases.

Question 6.
The vertices of a ∆ ABC are A(4, 6), B(l, 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 A ADE and compare it with the area of ∆ ABC.
Solution:
Using only Coordinate Geometry

Since D divides AB in the ratio 1:3, therefore coordinates of D are
\(\left(\frac{1 \times 1+3 \times 4}{1+3}, \frac{1 \times 5+3 \times 6}{4}\right)\), i.e., (\(\frac { 13 }{ 4 }\), \(\frac { 23 }{ 4 }\))
From \(\frac { AE }{ AC }\) = \(\frac { 1 }{4 }\) , we find that
\(\frac { AE }{ AC }\) = \(\frac { 1 }{ 3 }\)
Since E divides AC in the ratio 1:3, therefore coordinates of E are

Also, we find the area (∆ ABC) :
Area ∆ ABC = \(\frac { 15 }{ 2 }\) sq. units [already found]
Area (∆ ADE) : Area (∆ ABC)
= \(\frac { 15 }{ 32 }\) : \(\frac { 15 }{ 2 }\) = \(\frac { 1 }{ 32 }\) : \(\frac { 1 }{ 2 }\) = 1 : 16

Aliter :
We know that if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

∴ DE || BC
Clearly, A ADE ~ A ABC (Being equiangular, so similar)
We know that the ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides.

Also, from (1), Area (∆ ADE) : Area (∆ ABC) = 1 : 16.

Question 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 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:
Let A(4, 2), B(6, 5) and C(1, 4) be the vertices of A ABC.
(i) Since AD is the median of ∆ ABC, therefore
D is the mid-point of BC. Its coordinates are

\(\left(\frac{6+1}{2}, \frac{5+4}{2}\right)\), i.e., \(\left(\frac{7}{2}, \frac{9}{2}\right)\)

(ii) Since P divides AD in the ratio 2 : 1, so its coordinates are

(iii) Since BE is the median of ∆ ABC, so E is the mid-point of AC and its coordinates are E\(\left(\frac{4+1}{2}, \frac{2+4}{2}\right)\), i.e., E(\(\frac { 5 }{ 2 }\), 3)

Since Q divides BE in the ratio 2 : 1, so its coordinates are

Since CF is the median of ∆ ABC, so F is the mid-point of AB. Therefore, its coordinates are F\(\left(\frac{4+6}{2}, \frac{2+5}{2}\right)\). i.e., F(5, \(\frac { 7 }{ 2 }\))
Since R divides CF in the ratio 2 : 1, so its coordinates are

Since R divides CF in the ratio 2 : 1, so its coordinates are

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

(v) Let A(x1, y1, x2, y2) and C(x3, y3 be the vertices of ∆ ABC whose medians are AD, BE and CF respectively. So, D, E and F are respectively the mid-points of BC, CA and AB.

The coordinates of E are \(\left(\frac{x_{1}+x_{3}}{2}, \frac{y_{1}+y_{3}}{2}\right)\) coordinates of a point dividing BE in the ratio 2 : 1 are

The coordinates of E are \(\left(\frac{x_{1}+x_{3}}{2}, \frac{y_{1}+y_{3}}{2}\right)\). 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

AD, BE and CF and divides them in the ratio 2 : 1.
∴ The medians of a triangle are concurrent and the coordinates of the centroid are
\(\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right)\)

Question 8.
ABCD is a rectangle formed by joining 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:
Various points are marked in the figure shown on the next page.

Since all the sides are equal but the diagonals are not equal, therefore PQRS is a rhombus.