**Download NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.3 – Coordinate Geometry**

**1. Find the area of the triangle whose vertices are:**

**i. (2, 3), (-1, 0), (2, -4)**

Area = |2×(0 – (-4)) + (-1)×(-4 – 3) + 2×(3 – 0)|

= |2×4 + (-1)×(-7) + 2×3|

= |8 + 7 + 6|

= sq. units

**ii. (-5, -1), (3, -5), (5, 2)**

Area = |(-5)×(-5 – 2) + 3×(2 – (-1)) + 5×(-1 – (-5))|

= |(-5)×(-7) + 3×3 + 5×4|

= |35 + 9 + 20|

=

= 32 sq. units

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

**i. (7, -2), (5, 1), (3, k)**

Area = |7×(1 – k) + 5×(k – (-2)) + 3×(-2 – 1)|

= |7 – 7k + 5k + 10 – 9|

= |8 – 2k|

= |4 – k|

For three points to be collinear, area of the triangle formed by the points should be zero

|4 – k|= 0

k = 4

**ii. (8, 1), (k, -4), (2, -5)**

Area = |8×(-4 – (-5)) + k×(-5 – 1) + 2×(1 – (-4))|

= |8×(1) + k×(-6) + 2×5|

= |8 + -6k + 10|

= |18 – 6k|

= |9 – 3k|

For three points to be collinear, area of the triangle formed by the points should be zero

|9 – 3k|= 0

k = 3

**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 A, B and C denote the points (0, -1), (2, 1) and (0, 3)

Let D, E and F be the mid-points of the sides AB, BC and CA respectively

∴ D = = (1, 0)

E = = (1, 2)

F = = (0, 1)

Area of △ABC = |0×(1 – 3) + 2×(3 – (-1)) + 0×(-1 – 1)|

= |0+2×4+0|

= 4 sq. units

Area of △DEF = |1×(2 – 1) + 1×(1 – 0) + 0×(0 – 2)|

= |1×1+1×1+0|

= 1 sq. unit

Ratio of area of △DEF to △ABC = = 0.25

**4. Find the area of the quadrilateral whose vertices, taken in order, are (-4, -2), (-3, -5), (3, -2) and (2, 3).**

Let A, B, C and D denote the points (-4, -2), (-3, -5), (3, -2) and (2, 3).

Area of ABCD = ar(△ABC) + ar(△ACD)

= |(-4)×(-5 – (-2)) + (-3)×(-2 – (-2)) + 3×(-2 – (-5))|+ |(-4)×(-2 – 3) + 3×(3 – (-2)) + 2×(-2 – (-2))|

= |(-4)×(-3) + (-3)×0 + 3×3| + |(-4)×(-5) + 3×5 + 2×0|

= |12 + 9| + |20 + 15|

= +

=

= 28 sq. units

**5. Verify that a median of a triangle divides it into two triangles of equal areas for the △ABC whose vertices are A(4, -6), B(3, -2) and C(5, 2).**

Let AD be one of the medians of △ABC

∴ D is the mid-point of BC

D = = (4, 0)

Area of △ABD = |4×(-2 – 0) + 3×(0 – (-6)) + 4×(-6 – (-2))|

= |4×(-2) + 3×6 + 4×(-4)|

= |-8 + 18 – 16|

=

= 3 sq. unit

Area of △ACD = |4×(2 – 0) + 5×(0 – (-6)) + 4×(-6 – 2)|

= |4×2 + 5×6 + 4×(-8)|

= |8 + 30 + -32|

=

= 3 sq. unit

∵ ar(△ABD) = ar(△ACD)

∴ Median AD divides the △ABC into two triangles of equal areas

Similarly, it can be verified for other two medians

Hence, a median of a triangle divides it into two triangles of equal areas.

**Download NCERT Solutions For Class 10 Maths Chapter 7 Exercise 7.3 – Coordinate Geometry**

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