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Exercise - 7.3

Quadratic Equations

**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)

(i) The area of the triangle formed by the vertices A(2, 3), B(–1, 0) and C(2, – 4) By using the area of triangle formula, is given by = 1/2[x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)] = 1/2[2(0 + 4) + (-1)(-4 - 3) + (2)(3 - 0)] = 1/2[8 + 7 + 6] = 1/2 x 21 = 21/2 So, the area of the triangle is 21/2 square units.

(ii) The area of the triangle formed by the vertices A(–5, –1), B(3, –5) and C(5, 2) By using the area of triangle formula, is given by = 1/2[x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)] = 1/2[(-5)(-5 - 2) + 3(2 + 1) + 5(-1 + 5)] = 1/2[35 + 9 + 20] = 1/2 x 64 = 32 So, the area of the triangle is 32 square units.

**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)

(i) Since the given points are collinear, the area of the triangle formed by them must be 0, i.e., 1/2[7(1 - k) + 5(k + 2) + 3(-2 - 1)] = 0 1/2[7 - 7k + 5k + 10 - 6 - 3] = 0 1/2[-2k + 8] = 0 -2k + 8 = 0 -2k = -8 k = 8/2 k = 4

(ii) Since the given points are collinear, the area of the triangle formed by them must be 0, i.e., 1/2[8(-4 + 5) + k(-5 - 1) + 2(1 + 4)] = 0 1/2[8 - 6k + 10] = 0 1/2[-6k + 18] = 0 -6k + 18 = 0 -6k = -18 k = 18/6 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.

Let the vertices of the triangle be A (0, −1), B (2, 1), C (0, 3). Let D, E, F be the mid-points of the sides of this triangle. Coordinates of D, E, and F are given by D = [(0+2)/2, (-1+1)/2] = (1, 0) E = [(0+0)/2, (3-1)/2] = (0, 1) F = [(2+0)/2, (1+3)/2] = (1, 2) By using the area of triangle formula, is given by = 1/2[x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)] Now, Area of triange DEF, = 1/2[1(1 - 2) + 0(2 - 0) + 1(0 - 1)] = 1/2[-1 + 0 - 1] = 1/2 x (-2) = -1 Since area is a measure, which cannot be negative, we will take the numerical value of – 1, i.e., 1. So, the area of the triangle DEF is 1 square units. Also, Area of triange ABC, = 1/2[0(1 - 3) + 2(3 + 1) + 0(-1 - 1)] = 1/2[0 + 8 + 0] = 1/2 x 8 = 4 So, the area of the triangle ABC is 4 square units. Therefore, required ratio is 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).

Let the vertices of the quadrilateral be A (−4, −2), B (−3, −5), C (3, −2), and D (2,3). Join AC to form two triangles ΔABC and ΔACD. By using the area of triangle formula, is given by = 1/2[x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)] Now, Area of triange ABC, = 1/2[(-4)(-5 + 2) + (-3)(-2 + 2) + 3(-2 + 5)] = 1/2[12 + 0 + 9] = 1/2 x 21 = 21/2 So, the area of the triangle ABC is 21/2 square units. Also, Area of triange ACD, = 1/2[(-4)(-2 - 3) + 3(3 + 2) + 2(-2 + 2)] = 1/2[20 + 15 + 0] = 1/2 x 35 = 35/2 So, the area of the triangle ACD is 35/2 square units. Area of quadrilateral ABCD = area of triangle ABC + area of triangle ACD = 21/2 + 35/2 = 56/2 = 28 square 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 Δ ABC whose vertices are A(4, – 6), B(3, –2) and C(5, 2).

Let the vertices of the triangle be A (4, −6), B (3, −2), and C (5, 2). Let D be the mid-point of side BC of ΔABC. Therefore, AD is the median in ΔABC. Coordinates of point D = [(3+5)/2, (-2+2)/2] = (4, 0) By using the area of triangle formula, is given by = 1/2[x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)] Now, Area of triange ABD, = 1/2[4(-2 - 0) + 3(0 + 6) + 4(-6 + 2)] = 1/2[-8 + 18 - 16] = 1/2 x (-6) = -3 Since area is a measure, which cannot be negative, we will take the numerical value of – 3, i.e., 3. So, the area of the triangle ABC is 3 square units. Also, Area of triange ACD, = 1/2[4(2 - 0) + 5(0 + 6) + 4(-6 - 2)] = 1/2[8 + 30 - 32] = 1/2 x 6 = 3 So, the area of the triangle ADC is 3 square units. Clearly, median AD has divided ΔABC in two triangles of equal areas.

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