Find the area of the triangle whose vertices are : (ii) (–5, –1), (3, –5), (5, 2)

 exercise 7.3 co ordinate geometry chapter 7 cbse ncert 10th mathematics

 

 Find the area of the triangle whose vertices are :

(ii) (–5, –1), (3, –5), (5, 2)

 

Area of triangle = (1/2) [ (-5)[-5-2] +3[2-(-1)]+5[-1-(-5)]

=(1/2)[35+9+20]=32 sq.units.

2. In each of the following find the value of ‘k’, for which the points are collinear.
 

 (7, –2), (5, 1), (3, k)

collinear means that area of the triangle formed by the points is zero

so 

(1/2) [ 7(1-k)+5[k-(-2)]+3[-2-1] ] = 0

(1/2) [7-7k+5k+10-9]=0

(1/2[8 -2k] = 0

8 -2k =0

2k=8

k=8/2

k=4

 

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ncert cbse 10th mathematics

 

co ordinate geometry chapter 7

exercise 7.4 optional exercise  

 

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 

 

2. Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.

solution 

 

3. Find the centre of a circle passing through the points (6, – 6), (3, – 7) and (3, 3).

solution

4. The two opposite vertices of a square are (–1, 2) and (3, 2). Find the coordinates of the other two vertices.

solution   

6. The vertices of a ∆ ABC are A(4, 6), B(1, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively so that [AD/AB] =[AE/AC] =[1/4] Calculate the area of  ∆ ADE and compare it with the area of ∆ ABC

solution

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  

solution 

 

8. ABCD is a rectangle formed by the points A(–1, –1), B(– 1, 4), C(5, 4) and

 D(5, – 1). P, Q, 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

 

exercise 7.3

 Find the area of the triangle whose vertices are

  (2, 3), (–1, 0), (2, – 4)

 solution

(ii) (–5, –1), (3, –5), (5, 2)

solution

 

2. In each of the following find the value of ‘k’, for which the points are collinear.
 

 (7, –2), (5, 1), (3, k)

 solution

 

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