TOPICS

Unit-7(Examples)

Coordinate Geometry

**Example-1 :-** Do the points (3, 2), (–2, –3) and (2, 3) form a triangle? If so, name the type of triangle formed.

Let us apply the distance formula to find the distances PQ, QR and PR, where P(3, 2), Q(–2, –3) and R(2, 3) are the given points. We have PQ = √(3+2)² + (2+3)² = √(5)² + (5)² = √50 QR = √(-2-2)² + (-3-3)² = √(-4)² + (-6)² = √52 PR = √(3-2)² + (2-3)² = √(1)² + (-1)² = √2 Since the sum of any two of these distances is greater than the third distance, therefore, the points P, Q and R form a triangle. Also, PQ² + PR² = QR², (√50)² + (√2)² = (√52)² 50 + 2 = 52 52 = 52 L.H.S = R.H.S by the converse of Pythagoras theorem, we have ∠ P = 90°. Therefore, PQR is a right triangle.

**Example-2 :-** Show that the points (1, 7), (4, 2), (–1, –1) and (– 4, 4) are the vertices of a square.

Let A(1, 7), B(4, 2), C(–1, –1) and D(– 4, 4) be the given points. One way of showing that ABCD is a square is to use the property that all its sides should be equal and both its digonals should also be equal. Now, AB = √(1-4)² + (7-2)² = √(3)² + (5)² = √34 BC = √(4+1)² + (2+1)² = √(5)² + (3)² = √34 CD = √(-1+4)² + (-1-4)² = √(3)² + (-5)² = √34 DA = √(1+4)² + (7-4)² = √(5)² + (3)² = √34 AC = √(1+1)² + (7+1)² = √(2)² + (8)² = √68 BD = √(4+4)² + (2-4)² = √(8)² + (-2)² = √68 Since, AB = BC = CD = DA and AC = BD, all the four sides of the quadrilateral ABCD are equal and its diagonals AC and BD are also equal. Thereore, ABCD is a square.

**Example-3 :-** Figure shows the arrangement of desks in a classroom. Ashima, Bharti and Camella are seated at A(3, 1), B(6, 4) and C(8, 6) respectively.
Do you think they are seated in a line? Give reasons for your answer.

Using the distance formula, we have AB = √(6-3)² + (4-1)² = √(3)² + (3)² = √18 = 3√2 BC = √(8-6)² + (6-4)² = √(2)² + (2)² = √8 = 2√2 AC = √(8-3)² + (6-1)² = √(5)² + (5)² = √50 = 5√2 Since, AB + BC = AC, 3√2 + 2√2 = 5√2 5√2 = 5√2 L.H.S = R.H.S So, we can say that the points A, B and C are collinear. Therefore, they are seated in a line.

**Example-4 :-** : Find a relation between x and y such that the point (x , y) is equidistant from the points (7, 1) and (3, 5).

Let P(x, y) be equidistant from the points A(7, 1) and B(3, 5). We are given that AP = BP. So, AP² = BP² i.e., (x – 7)² + (y – 1)² =( x – 3)² + (y – 5)² i.e., x² – 14x + 49 + y² – 2y + 1 = x² – 6x + 9 + y² – 10y + 25 i.e., x – y = 2 which is the required relation.

**Example-5 :-** Find a point on the y-axis which is equidistant from the points A(6, 5) and B(– 4, 3).

We know that a point on the y-axis is of the form (0, y). So, let the point P(0, y) be equidistant from A and B. We are given that AP = BP. So, AP² = BP² Then (6 – 0)² + (5 – y)² = (– 4 – 0)² + (3 – y)² i.e., 36 + 25 + y² – 10y = 16 + 9 + y² – 6y i.e., 4y = 36 i.e., y = 9 So, the required point is (0, 9).

**Example-6 :-** Find the coordinates of the point which divides the line segment joining the points (4, – 3) and (8, 5) in the ratio 3 : 1 internally.

Let P(x, y) be the required point. Given that : x₁ = 4, y₁ = -3 ; x₂ = 8, y₂ = 5 ; m₁ = 3, m₂ = 1 Using the section formula, we get x = [(3x8 + 1x4)/(3 + 1)] = (24 + 4)/4 = 28/4 = 7 y = [(3x5 + 1x(-3))/(3 + 1)] = (15 - 3)/4 = 12/4 = 3 Therefore, (7, 3) is the required point.

**Example-7 :-** In what ratio does the point (– 4, 6) divide the line segment joining the points A(– 6, 10) and B(3, – 8)?

Let (– 4, 6) divide AB internally in the ratio m1 : m2. Using the section formula, we get (– 4, 6) = [(3m₁ - 6m₂)/(m₁ + m₂), (-8m₁ + 10m₂)/(m₁ + m₂)] -4 = (3m₁ - 6m₂)/(m₁ + m₂) -4m₁ - 4m₂ = 3m₁ + 6m₂ -4m₁ - 3m₁ = -6m₂ + 4m₂ -7m₁ = -2m₂ m₁/m₂ = 2/7 m₁:m₂ = 2:7 Therefore, the point (– 4, 6) divides the line segment joining the points A(– 6, 10) and B(3, – 8) in the ratio 2 : 7.

**Example-8 :-** Find the coordinates of the points of trisection (i.e., points dividing in three equal parts) of the line segment joining the points A(2, – 2) and B(– 7, 4).

Let P and Q be the points of trisection of AB i.e., AP = PQ = QB Therefore, P divides AB internally in the ratio 1 : 2. Therefore, the coordinates of P, by applying the section formula, are [(1x(-7) + 2x2)/(1 + 2), (1x4 + 2x(-2))/(1 + 2)] i.e., (-1, 0) Now, Q also divides AB internally in the ratio 2 : 1. So, the coordinates of Q are [(2x(-7) + 1x2)/(2 + 1), (2x4 + 1x(-2))/(2 + 1)] i.e., (-4, 2) Therefore, the coordinates of the points of trisection of the line segment joining A and B are (–1, 0) and (– 4, 2).

**Example-9 :-** Find the ratio in which the y-axis divides the line segment joining the points (5, – 6) and (–1, – 4). Also find the point of intersection.

Let the ratio be k : 1. Then by the section formula, the coordinates of the point which divides AB in the ratio k : 1 are [(-k + 5)/(k + 1), (-4k - 6)/(k + 1)] This point lies on the y-axis, and we know that on the y-axis the abscissa is 0. Therefore, (-k + 5)/(k + 1) = 0 -k + 5 = 0 -k = -5 k = 5 i.e., the ratio is 5 : 1. Putting the value of k = 5, we get the point of intersection as (0, -13/3).

**Example-10 :-** If the points A(6, 1), B(8, 2), C(9, 4) and D(p, 3) are the vertices of a parallelogram, taken in order, find the value of p.

We know that diagonals of a parallelogram bisect each other. So, the coordinates of the mid-point of AC = coordinates of the mid-point of BD [(6 + 9)/2, (1 + 4)/2] = [(8 + p)/2, (2 + 3)/2] (15/2, 5/2) = [(8 + p)/2, 5/2] 15/2 = (8 + p)/2 15 = 8 + p p = 15 - 8 p = 7

**Example-11 :-** Find the area of a triangle whose vertices are (1, –1), (– 4, 6) and (–3, –5).

The area of the triangle formed by the vertices A(1, –1), B(– 4, 6) and C (–3, –5), By using the area of triangle formula, is given by = 1/2[x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)] = 1/2[1(6 + 5) + (-4)(-5 + 1) + (-3)(-1 - 6)] = 1/2[11 + 16 + 21] = 1/2 x 48 = 24 So, the area of the triangle is 24 square units.

**Example-12 :-** Find the area of a triangle formed by the points A(5, 2), B(4, 7) and C (7, – 4).

The area of the triangle formed by the vertices A(5, 2), B(4, 7) and C (7, – 4), By using the area of triangle formula, is given by = 1/2[x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)] = 1/2[5(7 + 4) + 4(-4 - 2) + 7(2 - 7)] = 1/2[55 - 24 - 35] = 1/2 x (-4) = -2 Since area is a measure, which cannot be negative, we will take the numerical value of – 2, i.e., 2. So, the area of the triangle is 2 square units.

**Example-13 :-** Find the area of the triangle formed by the points P(–1.5, 3), Q(6, –2) and R(–3, 4).

The area of the triangle formed by the vertices P(–1.5, 3), Q(6, –2) and R(–3, 4), By using the area of triangle formula, is given by = 1/2[x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)] = 1/2[(-1.5)(-2 - 4) + 6(4 - 3) + (-3)(3 + 2)] = 1/2[9.0 + 6 - 15] = 1/2 x 0 = 0 If the area of a triangle is 0 square units, then its vertices will be collinear.

**Example-14 :-** Find the value of k if the points A(2, 3), B(4, k) and C(6, –3) are collinear.

Since the given points are collinear, the area of the triangle formed by them must be 0, i.e., 1/2[2(k + 3) + 4(-3 - 3) + 6(3 - k)] = 0 1/2[2k + 6 - 24 + 18 - 6k] = 0 1/2[-4k + 0] = 0 -4k = 0 k = 0

**Example-15 :-** If A(–5, 7), B(– 4, –5), C(–1, –6) and D(4, 5) are the vertices of a quadrilateral, find the area of the quadrilateral ABCD.

By joining B to D, you will get two triangles ABD and BCD. Now, the area of Δ ABD = 1/2[(-5)(-5 - 5) + (-4)(5 - 7) + 4(7 + 5)] = 1/2[50 + 8 + 48] = 1/2 x 106 = 53 square units Also, the area of Δ BCD = 1/2[(-4)(-6 - 5) + (-1)(5 + 5) + 4(-5 + 6)] = 1/2[44 - 10 + 4] = 1/2 x 38 = 19 square units So, the area of quadrilateral ABCD = 53 + 19 = 72 square units.

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