Coordinate Geometry NCERT Exercise 7.3 Class Ten Mathematics

Coordinate Geometry

Exercise 7.3 NCERT

Question 1: Find the area of the triangle whose vertices are:

  1. (2, 3), (-1, 0), (2, 4)
  2. (-5, -1), (3, -5), (5, 2)

Solution: (a) We have; `x_1 = 2`, `y_1 = 3`, `x_2 = -1`, `y_2 = 0`, `x_3 =2`, `y_3 = 4`

Area of triangle can be calculated as follows:

`= ½ (x_1(y_2 – y_3) + x_2 (y_3 – y_1) + x_3 (y_1 – y_2))`

`=1/2 (2(0 + 4) + (-1)(- 4 – 3) + 2(3 – 0))`

`= ½ (8 + 7 + 6)`

`= ½ xx 21 = 21/2` sq unit

Solution: (b) We have; `x_1 = -5`, `y_1 = -1`, `x_2 = 3`, `y_2 = -5`, `x_3 = 5`, `y_3 = 2`

Area of triangle can be calculated as follow:

`= ½ (x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2))`

`= ½ ( - 5(-5 – 2) + 3(2 + 1) + 5(-1 +5))`

`= ½ (35 + 9 + 20)`

`= ½ xx 64 = 32` sq unit


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

(a) (7, -2), (5, 1), (3, k)

Solution: Since the points are collinear, the area of triangle formed by them must be zero.

Or, `½ (x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2)) = 0`

Or, `½ (7(1 – k) + 5(k + 2) + 3( - 2 – 1)) = 0`

Or, `7 – 7k + 5k + 10 – 9 = 0`

Or, `8 – 2k = 0`

Or, `2k = 8`

Or `k = 4`

(b) (8, 1), (k, -4), (2, -5)

Solution: Since the points are collinear, the area of triangle formed by them must be zero.

Or, `½ (x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2)) = 0`

Or, `½ (8( - 4 + 5) + k( - 5 – 1) + 2(1 + 4)) = 0`

Or, `8 – 6k + 10 = 0`

Or, `18 – 6k = 0`

Or, `6k = 18`

Or, `k = 3`


Question 3: Find the area of the triangle formed by joining the midpoints 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.

Solution: A = (0, -1), B = (2, 1), C = (0, 3)

Coordinates of midpoint of AB can be calculated as follows:

`x=(0+2)/(2)=1`

`y=(-1+1)/(2)=0`

Coordinates of midpoint of BC:

`x=(0+0)/(2)=0`

`y=(1+3)/(2)=2`

Coordinates of midpoint of AC:

`x=(0+0)/(2)=0`

`y=(-1+3)/(2)=1`

We have; D = (1, 0), E = (1, 2), F = (0, 1)

Area of triangle DEF:

`= ½ (x_1(y_2 – y_3) + x_2 (y_3 – y_1) + x_3 (y_1 – y_2))`

`= ½ (1(2 – 1) + 1 (1 – 0) + 0(0 – 2))`

`= ½ (1 + 1) = 1` sq unit

Area of triangle ABC:

`= ½ (x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1– y_2))`

`= ½ (0(1 – 3) + 2(3 + 1) + 0( - 1 – 1))`

`= ½ (0 + 8 + 0 ) = 4` sq unit

Ratio of areas = 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).

Solution: Area of quadrilateral ABCD = Area ABD + Area BCD

Area of ABD:

`= ½ (x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2))`

`= ½ (- 4(-5 – 3) + - 3(3 + 2) + 2(-2 + 5))`

`= ½ (32 – 15 + 6)`

`= ½ xx 23 = 23/2` sq unit

Area of BCD:

`= ½ (x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3 (y_1 – y_2))`

`= ½ (-3(-2 – 3) + 3(3 + 5) + 2 (- 5 + 2))`

`= ½ (15 + 24 – 6)`

`= ½ xx 33 = 33/2` sq unit

Hence, area of ABCD `= 23/2 + 33/2 = 28` sq unit

Question 5: You have studied in class IX, that median of a triangle divides it into two triangles of equal areas. Verify this result for triangle ABC whose vertices are; A (4, -6), B (3,-2) and C (5, 2).

Solution: Area of triangle ABC can be calculated as follows:

`= ½ (x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2))`

`= ½ (4(-2 – 2) + 3(2 + 6) + 5( - 6 + 2))`

`= ½ ( - 16 + 24 – 20)`

`= ½ xx (– 12) = - 6` sq unit

Let us now assume median AD where D bisects AC. Coordinates of D can be calculated as follows:

`x=(4+5)/(2)=9/2`

`y=(-6+2)/(2)=-2`

Area of triangle ABD can be calculated as follows:

`= ½ (x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2))`

`= ½ (4 (-2 + 2) + 3 (- 2 + 6) + 9/2 (-6 + 2))`

`= ½ (0 + 12 – 18)`

`= ½ xx (-6) = - 3` sq unit

This shows that area of triangle ABD is half the area of triangle ABC and proves that a median divides a triangle into two triangles of equal areas.



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