NCERT Exemplar: Coordinate Geometry - 2 | Mathematics (Maths) Class 10 PDF Download
EXERCISE 7.3
Q.1. Name the type of triangle formed by the points A (-5, 6), B (-4, -2) and C (7, 5).
Given the points A(-5, 6), B(-4, -2) and C(7, 5). We find the three side lengths using the distance formula.
Formula: The distance between two points (x1 , y1) and (x2 , y2) is
√[(x2 - x1)2 + (y2 - y1)2]Distance AB = √[(-4 - (-5))2 + (-2 - 6)2]
= √[(1)2 + (-8)2]
= √(1 + 64)
= √65Distance BC = √[(7 - (-4))2 + (5 - (-2))2]
= √[(11)2 + (7)2]
= √(121 + 49)
= √170Distance AC = √[(7 - (-5))2 + (5 - 6)2]
= √[(12)2 + (-1)2]
= √(144 + 1)
= √145All three side lengths √65, √170 and √145 are different, so none of the sides are equal. Hence triangle ABC is a scalene triangle.
Q.2. Find the points on the x-axis which are at a distance of 2√5 from the point (7, -4). How many such points are there?
Let A(x, 0) be a point on the x-axis at distance 2√5 from B(7, -4).
Formula: The distance between two points P (x1 , y1) and Q (x2 , y2) is
√[(x2 - x1)2 + (y2 - y1)2]Distance AB = √[(7 - x)2 + (-4 - 0)2] = 2√5
On squaring both sides,
(7 - x)2 + 16 = 20
By using algebraic identity,
(a - b)2 = a2 - 2ab + b2
So, (7 - x)2 = (7)2 - 2(7)(x) + (x)²
= 49 - 14x + x2
49 - 14x + x2 + 16= 20
x2 - 14x + 49 - 20 + 16 = 0
x2 - 14x + 45 = 0
x2 - 9x - 5x + 45 = 0
x(x - 9) - 5(x - 9) = 0(x - 5)(x - 9) = 0
Now, x - 5 = 0
x = 5
Also, x - 9 =0
x = 9
Therefore, the points (5, 0) and (9, 0) are at a distance of 2√5 from the point (7, -4).
Q.3. What type of a quadrilateral do the points A (2, -2), B (7, 3), C (11, -1) and D (6, -6) taken in that order, form?
Calculate the side lengths and diagonals.
Formula: The distance between two points P (x1 , y1) and Q (x2 , y2) is
√[(x2 - x1)2 + (y2 - y1)2]AB = √[(7 - 2)2 + (3 - (-2))2] = √(52 + 52) = 5√2
BC = √[(11 - 7)2 + (-1 - 3)2] = √(42 + (-4)2) = 4√2
CD = √[(6 - 11)2 + (-6 - (-1))2] = √((-5)2 + (-5)2) = 5√2
DA = √[(2 - 6)2 + (-2 - (-6))2] = √((-4)2 + 42) = 4√2
Diagonals:
AC = √[(11 - 2)2 + (-1 - (-2))2] = √(92 + 12) = √82
BD = √[(6 - 7)2 + (-6 - 3)2] = √((-1)2 + (-9)2) = √82Opposite sides are equal (AB = CD and BC = DA) and the diagonals are equal (AC = BD). These properties imply the quadrilateral is a rectangle.
Q.4. Find the value of a, if the distance between the points A (-3, -14) and B (a, -5) is 9 units.
Formula: The distance between two points P (x1 , y1) and Q (x2 , y2) is
√[(x2 - x1)2 + (y2 - y1)2]Distance AB = √[(a - (-3))2 + (-5 - (-14))2] = 9
√[(a + 3)2 + 92] = 9
Square both sides:
(a + 3)2 + 81 = 81(a + 3)2 = 0
Hence a + 3 = 0 → a = -3.
Q.5. Find a point which is equidistant from the points A (-5, 4) and B (-1, 6)? How many such points are there?
Formula: The distance between two points P (x1 , y1) and Q (x2 , y2) is
√[(x2 - x1)2 + (y2 - y1)2]Let P(x, y) satisfy PA = PB.
√[(x + 5)2 + (y - 4)2] = √[(x + 1)2 + (y - 6)2]
Square both sides and simplify:
(x + 5)2 + (y - 4)2 = (x + 1)2 + (y - 6)2Expand and cancel common terms to obtain the equation of the perpendicular bisector:
2x + y + 1 = 0The midpoint of AB is [ (-5 + (-1))/2, (4 + 6)/2 ] = (-3, 5), which satisfies 2(-3) + 5 + 1 = 0, so it lies on this line.
The locus of points equidistant from A and B is the straight line 2x + y + 1 = 0, so there are infinitely many such points. One example is the midpoint (-3, 5).
Q.6. Find the coordinates of the point Q on the x-axis which lies on the perpendicular bisector of the line segment joining the points A (-5, -2) and B(4, -2). Name the type of triangle formed by the points Q, A and B.
Let Q(x, 0). For Q to lie on the perpendicular bisector of AB we must have QA = QB.
Formula: The distance between two points P (x1 , y1) and Q (x2 , y2) is
√[(x2 - x1)2 + (y2 - y1)2]QA = √[(x + 5)2 + (0 + 2)2] = √[(x + 5)2 + 4]
QB = √[(x - 4)2 + (0 + 2)2] = √[(x - 4)2 + 4]
Equate and square:
(x + 5)2 + 4 = (x - 4)2 + 4 → (x + 5)2 = (x - 4)2Expand:
x2 + 10x + 25 = x2 - 8x + 1618x + 9 = 0 → x = -9/18 = -1/2
So Q = (-1/2, 0).
Compute QA and QB to check equality (optional): QA = QB = √[(4.5)2 + 4] = √24.25, so QA = QB.
Since QA = QB, triangle QAB has two equal sides and is therefore isosceles.
Q.7. Find the value of m if the points (5, 1), (-2, -3) and (8, 2m ) are collinear
Use the area condition for collinearity: area of triangle formed by the three points = 0.
The area of a triangle with vertices A (x1 , y1) , B (x2 , y2) and C (x3 , y3) is
1/2[x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)]Area = 1/2[5(-3 - 2m) + (-2)(2m - 1) + 8(1 - (-3))] = 0
Simplify:
5(-3 - 2m) + (-2)(2m - 1) + 8(4) = 0-15 - 10m - 4m + 2 + 32 = 0
-14m + 19 = 0 → 14m = 19 → m = 19/14.
Q.8. If the point A (2, -4) is equidistant from P (3, 8) and Q (-10, 3)), find the values of y. Also find distance PQ.
Given A(2, -4), P(3, 8) and Q(-10, y). Since A is equidistant from P and Q, AP = AQ.
Formula: The distance between two points P (x1 , y1) and Q (x2 , y2) is
√[(x2 - x1)2 + (y2 - y1)2]AP = √[(3 - 2)2 + (8 + 4)2] = √[1 + 144] = √145.
AQ = √[(-10 - 2)2 + (y + 4)2] = √[144 + (y + 4)2].
Equate and square:
144 + (y + 4)2 = 145 → (y + 4)2 = 1y + 4 = ±1
y = -3 or y = -5.
Now compute PQ for both values of y.
For y = -3: PQ = √[(3 - (-10))2 + (8 - (-3))2]
= √[(13)2 + (11)2]
= √(169 + 121)
= √290.For y = -5: PQ = √[(3 - (-10))2 + (8 - (-5))2]
= √[(13)2 + (13)2]
= √(169 + 169)
= √338 = 13√2.Thus y = -3 or -5, and the corresponding distances PQ are √290 and 13√2 respectively.
Q.9. Find the area of the triangle whose vertices are (-8, 4), (-6, 6) and (-3, 9).
Use the determinant formula (area expression):
Area = 1/2[ x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂) ]
= 1/2[ (-8)(6 - 9) + (-6)(9 - 4) + (-3)(4 - 6) ]
= 1/2[ (-8)(-3) + (-6)(5) + (-3)(-2) ]
= 1/2[ 24 - 30 + 6 ] = 1/2[0] = 0.
The area is 0, so the three points are collinear.
Q.10. In what ratio does the x-axis divide the line segment joining the points (- 4, - 6) and (-1, 7)? Find the coordinates of the point of division.
The coordinates of the point P(x, y) which divides the line segment joining the points A (x1 , y1) and B (x2 , y2) internally in the ratio k : 1 are [(kx2 + x1)/(k + 1) , (ky2+ y1)/(k + 1)]
Let the division ratio be k : 1 so that the point P is [(-k - 4)/(k + 1), (7k - 6)/(k + 1)].
P lies on the x-axis so y = 0 → (7k - 6)/(k + 1) = 0 → 7k - 6 = 0 → k = 6/7.
Thus the ratio is 6 : 7 (internally).
Now x-coordinate = [6(-1) + 7(-4)]/(6 + 7) = (-6 - 28)/13 = -34/13.
So the point of division is (-34/13, 0).
Q.11. Find the ratio in which the point 
divides the line segment joining the points 
and B (2, -5).
Given P(3/4, 5/12), A(1/2, 3/2) and B(2, -5). Let P divide AB internally in the ratio k : 1.
The coordinates of the point P(x, y) which divides the line segment joining the points A (x1 , y1) and B (x2 , y2) internally in the ratio k : 1 are [(kx2 + x1)/(k + 1), (ky2 + y1)/(k + 1)]For x-coordinate:
(2k + 1/2)/(k + 1) = 3/4 → 2k + 1/2 = 3/4(k + 1)Solve: (4k + 1)/2 = (3k + 3)/4 → 2(4k + 1) = 3k + 3 → 8k + 2 = 3k + 3 → 5k = 1 → k = 1/5.
For y-coordinate you obtain the same k after simplification. Hence the ratio is 1 : 5.
Q.12. If P (9a - 2, -b) divides line segment joining A (3a + 1, -3) and B (8a, 5) in the ratio 3 : 1, find the values of a and b.
The coordinates of the point P which divides the line segment joining the points A (x₁ , y₁) and B (x2 , y2) internally in the ratio m₁ : m₂ are
[(m1x2 + m2x1)/(m1 + m2) , (m1y2 + m2y1)/(m1 + m2)]Coordinates of P using section formula with ratio 3 : 1 are:
x-coordinate = [3·(8a) + 1·(3a + 1)]/4 = (24a + 3a + 1)/4 = (27a + 1)/4
Set equal to given x of P: (27a + 1)/4 = 9a - 2
27a + 1 = 36a - 8 → 9a = 9 → a = 1.
y-coordinate = [3·5 + 1·(-3)]/4 = (15 - 3)/4 = 12/4 = 3.
Given y of P is -b, so -b = 3 → b = -3.
Therefore a = 1 and b = -3.
Q.13. If (a, b) is the mid-point of the line segment joining the points A (10, -6) and B (k, 4) and a - 2b = 18, find the value of k and the distance AB.
Midpoint (a, b) = ( (10 + k)/2, (-6 + 4)/2 ) = ( (10 + k)/2, -1 ).
So b = -1. From a - 2b = 18 → a - 2(-1) = 18 → a + 2 = 18 → a = 16.
(10 + k)/2 = a = 16 → 10 + k = 32 → k = 22.
Distance AB = √[(22 - 10)2 + (4 - (-6))2] = √(122 + 102) = √(144 + 100) = √244 = 2√61.
Q.14. The centre of a circle is (2a, a - 7). Find the values of a if the circle passes through the point (11, -9) and has diameter 10√2 units.
Radius r = (diameter)/2 = 5√2.
The distance between two points P (x1 , y1) and Q (x2 , y2) is √[(x2 - x1)2 + (y2 - y1)2]Distance from centre (2a, a - 7) to (11, -9) equals r:
√[(11 - 2a)2 + (-9 - (a - 7))2] = 5√2
Square both sides:
(11 - 2a)2 + (-a - 2)2 = 50Expand:
121 - 44a + 4a2 + a2 + 4a + 4 = 505a2 - 40a + 125 - 50 = 0 → 5a2 - 40a + 75 = 0
Divide by 5: a2 - 8a + 15 = 0 → (a - 3)(a - 5) = 0
Therefore a = 3 or a = 5.
Q.15. The line segment joining the points A (3, 2) and B (5,1) is divided at the point P in
the ratio 1:2 and it lies on the line 3x - 18y + k = 0. Find the value of k.
P dividing AB in the ratio 1 : 2 has coordinates:
The coordinates of the point P which divides the line segment joining the points A (x₁ , y₁) and B (x2 , y2) internally in the ratio m₁ : m₂ are
[(m₁x2+m2x1)/(m₁+m2) , (m1y2+m2y1)/(m1+m2)]P = [ (1·5 + 2·3)/(1 + 2), (1·1 + 2·2)/(1 + 2) ] = (11/3, 5/3 ).
P lies on 3x - 18y + k = 0 → 3(11/3) - 18(5/3) + k = 0
11 - 6·5 + k = 0 → 11 - 30 + k = 0 → k = 19.
Q.16. If 
E (7, 3) and 
are the midpoints of sides of ∆ ABC, find the area of the ∆ ABC.
Given midpoints D(-1/2, 5/2), E(7, 3) and F(7/2, 7/2). Let A(x₁, y₁), B(x₂, y₂), C(x₃, y₃).
The coordinates of the mid-point of the line segment joining the points P (x1 , y1) and Q (x2 , y2) are [(x1 + x2)/2, (y1 + y2)/2]From midpoints:
(x₁ + x₂)/2 = -1/2 → x₁ + x₂ = -1
(y₁ + y₂)/2 = 5/2 → y₁ + y₂ = 5
(x₂ + x₃)/2 = 7 → x₂ + x₃ = 14
(y₂ + y₃)/2 = 3 → y₂ + y₃ = 6
(x₁ + x₃)/2 = 7/2 → x₁ + x₃ = 7
(y₁ + y₃)/2 = 7/2 → y₁ + y₃ = 7
Add the three x-equations: 2(x₁ + x₂ + x₃) = -1 + 14 + 7 = 20 → x₁ + x₂ + x₃ = 10.
Use x₁ + x₂ = -1 → x₃ = 11. Then x₁ = -4 and x₂ = 3.
Add the three y-equations: 2(y₁ + y₂ + y₃) = 5 + 6 + 7 = 18 → y₁ + y₂ + y₃ = 9.
Use y₁ + y₂ = 5 → y₃ = 4. Then y₁ = 3 and y₂ = 2.
So A(-4, 3), B(3, 2), C(11, 4).
Area = 1/2[ x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂) ]
= 1/2[ (-4)(2 - 4) + 3(4 - 3) + 11(3 - 2) ]
= 1/2[ (-4)(-2) + 3(1) + 11(1) ] = 1/2[ 8 + 3 + 11 ] = 1/2[22] = 11 square units.
Q.17. The points A (2, 9), B (a, 5) and C (5, 5) are the vertices of a triangle ABC right angled at B. Find the values of a and hence the area of ∆ABC.
Since the triangle ABC is a right triangle with B at right angle.
By pythagoras theorem,
AC2 = AB2 + BC2
The distance between two points P (x1 , y1) and Q (x2 , y2) is
√[(x2 - x1)2 + (y2 - y1)2]Since ∠B = 90°, AB² + BC² = AC².
AC = √[(5 - 2)2 + (5 - 9)2] = √(32 + (-4)2) = 5.
AB = √[(a - 2)2 + (5 - 9)2] = √[(a - 2)2 + 16].
BC = |5 - a|.
So 25 = (a - 2)2 + 16 + (5 - a)2.
Expand: 25 = a2 - 4a + 4 + 16 + 25 - 10a + a2
25 = 2a2 - 14a + 45 → 2a2 - 14a + 20 = 0 → a2 - 7a + 10 = 0
(a - 2)(a - 5) = 0 → a = 2 or 5.
a = 5 would make B coincide with C, so a = 2 is the required value (distinct vertices).
Vertices are A(2, 9), B(2, 5), C(5, 5). Taking BC as base of length 3 and height AB = 4, area = 1/2 × 3 × 4 = 6 square units.
Q.18. Find the coordinates of the point R on the line segment joining the points
P (-1, 3) and Q (2, 5) such that PR = 3/5PQ.
PR = 3/5 PQ implies PR : QR = 3 : 2, so R divides PQ internally in the ratio 3 : 2.
The coordinates of the point P which divides the line segment joining the points A (x1 , y1) and B (x2 , y2) internally in the ratio m1 : m2 are
[(m1x2 + m2x1)/m1 + m2 , (m1y2 + m2y1)/m1 + m2]Using section formula for P(-1, 3) and Q(2, 5) with m₁ : m₂ = 3 : 2:
R = [ (3·2 + 2·(-1))/5 , (3·5 + 2·3)/5 ] = [ (6 - 2)/5 , (15 + 6)/5 ] = (4/5, 21/5 ).
Q.19. Find the values of k if the points A (k + 1, 2k), B(3k, 2k + 3) and C(5k - 1, 5k) are collinear.
The area of a triangle with vertices A (x1 , y1) , B (x2 , y2) and C (x3 , y3) is
1/2[x1(y2 - y3) + x2(y3 - y1) + x₃(y1 - y₂)]
Area of triangle ABC = 0 gives:1/2[ (k + 1)(2k + 3 - 5k) + 3k(5k - 2k) + (5k - 1)(2k - (2k + 3)) ] = 0
Simplify:
(k + 1)(3 - 3k) + 3k(3k) + (5k - 1)(-3) = 0Expand and collect terms:
-3k² + 3k + 3 - 3k + 9k² - 15k + 3 = 0 → 6k² - 15k + 6 = 0Divide by 1 (or 3): 2k² - 5k + 2 = 0 → (2k - 1)(k - 2) = 0
Hence k = 1/2 or k = 2.
Q.20. Find the ratio in which the line 2x + 3y - 5 = 0 divides the line segment joining the points (8, -9) and (2, 1). Also find the coordinates of the point of division.
The coordinates of the point P(x, y) which divides the line segment joining the points A (x1 , y1) and B (x1 , y1) internally in the ratio k : 1 are [(kx2 + x1)/k +1, (ky2 + y1)/k + 1]
Let the point of division be P = ( (2k + 8)/(k + 1), (k - 9)/(k + 1) ) for ratio k : 1.
P lies on 2x + 3y - 5 = 0.
Substitute:2(2k + 8)/(k + 1) + 3(k - 9)/(k + 1) - 5 = 0 → 2(2k + 8) + 3(k - 9) = 5(k + 1)
4k + 16 + 3k - 27 = 5k + 5 → 7k - 11 = 5k + 5 → 2k = 16 → k = 8.
Thus the line divides the segment in the ratio 8 : 1 internally.
Coordinates of P: x = (2·8 + 8)/(8 + 1) = 24/9 = 8/3, y = (8 - 9)/(8 + 1) = -1/9.
So the point of division is (8/3, -1/9).
EXERCISE 7.4
Q.1. If (- 4, 3) and (4, 3) are two vertices of an equilateral triangle, find the coordinates of the third vertex, given that the origin lies in the interior of the triangle.
Given, (-4, 3) and (4, 3) are the two vertices of an equilateral triangle.
Also, the origin lies in the interior of the triangle.
We have to find the coordinates of the third vertex.
Let an equilateral triangle be ABC.
The vertices of the triangle be A(-4, 3) B(4, 3) and the third vertex is (x, y).
An equilateral triangle is a triangle in which all three sides have the same length.
An equilateral triangle is also equiangular as all three internal angles are congruent to each other and are each 60°.
So, AB = BC = AC
Considering AB = BC
Formula: The distance between two points P (x1 , y1) and Q (x2 , y2) is
√[(x2 - x1)2 + (y2 - y1)2]
Distance between A(-4, 3) and B(4, 3) = √[(4 - (-4))² + (3 - 3)²]
= √[(8)² + 0]
= √64
= 8
Distance between B(4, 3) and C(x, y) = √[(x - 4)² + (y - 3)²]
8 = √[(x - 4)² + (y - 3)²]
On squaring both sides,
64 = (x - 4)² + (y - 3)²
By using algebraic identity,
(a - b)² = a² - 2ab + b²
So, (x - 4)² = x² - 8x + 16
(y - 3)² = y² - 6y + 9
Now, 64 = x² - 8x + 16 + y² - 6y + 9
x² + y² - 8x - 6y = 64 - 25
x² + y² - 8x - 6y = 39 ------------------ (1)
Considering AB = AC,
Distance between A(-4, 3) and C(x, y) = √[(x - (-4))² + (y - 3)²]
= √[(x + 4)² + (y - 3)²]
Now, 8 = √[(x + 4)² + (y - 3)²]
On squaring both sides,
64 = (x + 4)² + (y - 3)²
By using algebraic identity,
(a + b)² = a² + 2ab + b²
So, (x + 4)² = x² + 8x + 16
64 = x² + 8x + 16 + y² - 6y + 9
x² + y² + 8x - 6y = 64 - 25
x² + y² + 8x - 6y = 39 ---------------- (2)
Subtracting (1) and (2),
x² + y² + 8x - 6y - (x² + y² - 8x - 6y) = 39 - 39
On simplification,
x² - x² + y² - y² + 8x - 6y + 8x + 6y = 0
8x + 8x = 0
16x = 0
x = 0
Put x = 0 in (1)
(0)² + y² - 8(0) - 6y = 39
y² - 6y - 39 = 0
Using the quadratic formula,
y = -b∙±√(b² - 4ac)/2a
Here, a = 1, b = -6 and c = -39
So, y = 6 ± √((-6)² - 4(1)(-39))/2(1)
= 6 ± √(36 + 156)/2
= (6 ± √192)/2
y = 3 ± 4√3
Now, y = 3 - 4√3
y = 3 + 4√3
Since the origin lies in the interior of the triangle, the x coordinate of the third vertex is zero.y = 3 + 4√3 lies outside the triangle. So, it is not possible.
y = 3 - 4√3 lies inside the triangle.
Therefore, the coordinates of the third vertex = (0, 3 - 4√3)
Q.2. A (6, 1), B (8, 2) and C (9, 4) are three vertices of a parallelogram ABCD. If E is the mid point of DC, find the area of ΔADE.
Let D = (x, y). In a parallelogram the diagonals bisect each other, so midpoint of AC equals midpoint of BD.
Midpoint of AC = ( (6 + 9)/2, (1 + 4)/2 ) = (15/2, 5/2 ).
Midpoint of BD = ( (8 + x)/2, (2 + y)/2 ). Equate to get:
(8 + x)/2 = 15/2 → 8 + x = 15 → x = 7.
(2 + y)/2 = 5/2 → 2 + y = 5 → y = 3.
Thus D = (7, 3).
Midpoint E of DC: E = ( (7 + 9)/2, (3 + 4)/2 ) = (8, 7/2 ).
The area of a triangle with vertices A (x1 , y1) , B (x2 , y2) and C (x3 , y3) is
1/2[x1(y2 - y3) + x2(y3 - y1) + x₃(y1 - y₂)]Now area of ΔADE with A(6,1), D(7,3), E(8,7/2):
Area = 1/2[ 6(3 - 7/2) + 7(7/2 - 1) + 8(1 - 3) ]
Simplify in fractions:
3 - 7/2 = -1/2 → 6(-1/2) = -3
7/2 - 1 = 5/2 → 7·5/2 = 35/2
1 - 3 = -2 → 8(-2) = -16Sum = -3 + 35/2 - 16 = (-38 + 35)/2 = -3/2
Area = 1/2 · |-3/2| = 3/4 square units.
Q.3. The points A (x1, y1), B (x2, y2) and C (x3, y3) are the vertices of ΔABC.
(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 are the coordinates of the centroid of the triangle ABC?
(i) D is midpoint of BC, so D = ( (x₂ + x₃)/2 , (y₂ + y₃)/2 ).
(ii) P divides AD in ratio 2 : 1 (from A). Using section formula with m₁ : m₂ = 2 : 1 on A(x₁, y₁) and D,
P = [ (2·( (x₂ + x₃)/2 ) + 1·x₁)/(2 + 1), (2·( (y₂ + y₃)/2 ) + 1·y₁)/(2 + 1) ]
Simplify to P = ( (x₁ + 2x₂ + 2x₃)/3 , (y₁ + 2y₂ + 2y₃)/3 ).
(iii) E is midpoint of AC: E = ( (x₁ + x₃)/2, (y₁ + y₃)/2 ). Q divides BE in ratio 2 : 1 from B, so
Q = [ (2·( (x₁ + x₃)/2 ) + 1·x₂ )/3 , (2·( (y₁ + y₃)/2 ) + 1·y₂)/3 ] = ( (x₁ + x₂ + x₃)/3 , (y₁ + y₂ + y₃)/3 ).
Similarly F is midpoint of AB and R dividing CF in ratio 2 : 1 gives the same coordinates:
R = ( (x₁ + x₂ + x₃)/3 , (y₁ + y₂ + y₃)/3 ).
(iv) The centroid (point of intersection of medians) is the common point found above, namely
C = ( (x₁ + x₂ + x₃)/3 , (y₁ + y₂ + y₃)/3 ).
Q.4. If the points A (1, -2), B (2, 3) C (a, 2) and D (- 4, -3) form a parallelogram, find the value of a and height of the parallelogram taking AB as base.
Step 1: Use the property of parallelogram
In a parallelogram, the diagonals bisect each other. So midpoint of AC = midpoint of BD.
Midpoint of AC
A(1, −2), C(a, 2)
MAC = ((1 + a)/2 , (−2 + 2)/2) = ((1 + a)/2 , 0)Midpoint of BD
B(2, 3), D(−4, −3)
MBD = ((2 − 4)/2 , (3 − 3)/2) = (−1 , 0)Equating midpoints:
(1 + a)/2 = −1
1 + a = −2
a = −3If the points A(1, −2), B(2, 3), C(a, 2) and D(−4, −3) form a parallelogram, find the value of a and the height of the parallelogram taking AB as base.
From the property of parallelogram, diagonals bisect each other, so a = −3.
So C = (−3, 2).
We take AB as base. To find height, we find the perpendicular distance of D from line AB.
Slope of AB = (3 − (−2)) / (2 − 1) = 5
Equation of line AB: y + 2 = 5(x − 1) y = 5x − 7 5x − y − 7 = 0
Point D = (−4, −3)
Height = |5(−4) + (−1)(−3) − 7| / √(5² + (−1)²) = |−20 + 3 − 7| / √26 = |−24| / √26 = 24 / √26
Length of AB = √((2 − 1)² + (3 + 2)²) = √(1 + 25) = √26
Area = base × height = √26 × (24 / √26) = 24
Final Answer: a = −3 Height = 24 / √26 units
Q.5. Students of a school are standing in rows and columns in their playground for a drill practice. A, B, C and D are the positions of four students as shown in figure. Is it possible to place Jaspal in the drill in such a way that he is equidistant from each of the four students A, B, C and D? If so, what should be his position?
Coordinates read from the diagram: A(3, 5), B(7, 9), C(11, 5), D(7, 1).
Formula: The distance between two points P (x1 , y1) and Q (x2 , y2) is
√[(x2 - x1)2 + (y2 - y1)2]So, ABCD will be either square or rhombus.
Now, Diagonal
and diagonal
⇒BD=8 units
∴ Diagonal AC = Diagonal BD
So, the given quadrilateral ABCD is square. The point which is equidistant from point A, B, C, D of a square ABCD will be at the intersecting point of diagonals and diagonals bisect each other.
Hence, the required point O equidistant from A, B, C, D is mid point of any diagonal =
Hence, the required point is (7, 5).
Q.6. Ayush starts walking from his house to office. Instead of going to the office directly, he goes to a bank first, from there to his daughter's school and then reaches the office. What is the extra distance travelled by Ayush in reaching his office? (Assume that all distances covered are in straight lines). If the house is situated at (2, 4), bank at (5, 8), school at (13, 14) and office at (13, 26) and coordinates are in km.
Calculate the three legs:
Formula: The distance between two points P (x1 , y1) and Q (x2 , y2) is
√[(x2 - x1)2 + (y2 - y1)2]House to Bank: √[(5 - 2)2 + (8 - 4)2] = √(32 + 42) = 5 km.
Bank to School: √[(13 - 5)2 + (14 - 8)2] = √(82 + 62) = 10 km.
School to Office: √[(13 - 13)2 + (26 - 14)2] = √(0 + 122) = 12 km.
Total distance travelled = 5 + 10 + 12 = 27 km.
Direct distance from House(2,4) to Office(13,26): √[(13 - 2)2 + (26 - 4)2] = √(112 + 222) = √605 ≈ 24.6 km.
Extra distance = 27 - √605 ≈ 27 - 24.6 = 2.4 km (approximately).
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