Students can Download Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.3 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

## Karnataka State Syllabus Class 10 Maths 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)

Answer:

i) Let A (2, 3) = (x_{1}, y_{1})

B (- 1, 0) = (x_{2}, y_{2})

C (2, – 4) = (x_{3}, y_{3}).

Area of the triangle from the given data:

ii) Let A (- 5, – 1) = (x_{1}, y_{1})

B (3, -5) = (x_{2}, y_{2})

C (5, 2) = (x_{3}, y_{3}).

Area of the triangle from the given data :

area of ∆ PQR = 32 Sq 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)

Answer:

i) Let A (7, – 2)= (x_{1}, y_{1})

B (5, 1) = (x_{2}, y_{2})

C (3, k) = (x_{3}, y_{3}).

Points are collinear, therefore the area of the triangle formed by these is zero (0).

Hence the given points are collinear for K = 4.

ii) Let A (8, 1) = (x_{1}, y_{1})

B (k, – 4) = (x_{2}, y_{2})

C (2, – 5) = (x_{3}, y_{3}).

Area of Triangle ABC = 0

∴ ABC is a straight line.

\(\frac{1}{1}\) [x_{1} (y_{2} – y_{3}) + x_{3} (y_{3} – y_{1}) + x_{3} (y_{1} – y_{2})] = 0 .

[8(- 4 +5) + K (- 5 – 1) + 2(1 + 4)] = 0

8 x 1 – 6K + 2(5) = 0

8 – 6K + 10 = 0

– 6K + 18 = O

6K = 18

K = 18/6 = 3

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.

Answer:

Let the mid-points of ∆ABC are P, Q, R, and also the mid-points of AB, BC, and AC.

Now P, Q, R be the midpoints of BC, CA, and AB, respectively Midpoint formula

area of (∆ABC) = 4 sq units

Ratio of area of ∆PQR : ∆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)

Area of ∆ ABC

area of ∆ ADC = \(\frac{35}{2}\)

area of quadrilateral ABCD = Area of ∆ABC + area of ∆ACD

\(=\frac{21}{2}+\frac{35}{2}=\frac{56}{2}=28\)

area of quadrilateral ABCD = 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).

Answer:

Since AD is the median of ∆ ABC. D is the midpoint of BC

Coordinates of D are

Since the area cannot be negative. Therefor area of ∆ ABD = 3 sq. units → (1)

Area of ∆ ADC

ar (∆ ADC) = 3 sq. units → (2)

Form eqn (1) & (2)

are of ∆ ABD = area of ∆ ADC