(i) (2, 3), (-1, 0), (2, -4)
(ii) (-5, -1), (3, -5), (5, 2)
Area of a triangle = $\dfrac{1}{2} × [x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})]$
(i) (2, 3), (-1, 0), (2, -4)
Here,
$x_{1} = 2, x_{2} = -1, x_{3} = 2, y_{1} = 3, y_{2} = 0$ and $y_{3} = -4$
Substitute all the values in the above formula, and we get
Area of triangle = $\dfrac{1}{2}$ [2 {0- (-4)} + (-1) {(-4) – (3)} + 2 (3 – 0)]
=$\dfrac{ 1}{2}$ {8 + 7 + 6}
= $\dfrac{21}{2}$
So, the area of the triangle is $\dfrac{21}{2}$ square units.
(ii) (-5, -1), (3, -5), (5, 2)
Here,
$x_{1} = -5, x_{2} = 3, x_{3} = 5, y_{1} = -1, y_{2} = -5$ and $y_{3} = 2$
Area of the triangle = $\dfrac{1}{2}$ [-5 { (-5)- (2)} + 3(2-(-1)) + 5{-1 – (-5)}]
=$\dfrac{ 1}{2}${35 + 9 + 20} = 32
Therefore, the area of the triangle is 32 square units.
(i) (7, -2), (5, 1), (3, -k)
(ii) (8, 1), (k, -4), (2, -5)
(i) (7, -2), (5, 1), (3, -k)
For collinear points, the area of the triangle formed by them is always zero.
Let points (7, -2), (5, 1) and (3, k) be vertices of a triangle.
Area of triangle = $\dfrac{1}{2}$ [7 { 1- k} + 5(k-(-2)) + 3{(-2) – 1}] = 0
7 – 7k + 5k +10 -9 = 0
-2k + 8 = 0
k = 4
(ii) (8, 1), (k, -4), (2, -5)
For collinear points, the area of the triangle formed by them is zero.
Therefore, for points (8, 1), (k, – 4), and (2, – 5), area = 0
$\dfrac{1}{2}$ [8 { -4- (-5)} + k{(-5)-(1)} + 2{1 -(-4)}] = 0
8 – 6k + 10 = 0
6k = 18
k = 3
Let the vertices of the triangle be A (0, -1), B (2, 1), and C (0, 3).
Let D, E, and F be the mid-points of the sides of this triangle.
Coordinates of D, E, and F are given by
D = (0+$\dfrac{2}{2}$, -1+$\dfrac{1}{2}$ ) = (1, 0)
E = ( 0+$\dfrac{0}{2}$, -1+$\dfrac{3}{2}$ ) = (0, 1)
F = ( 0+$\dfrac{2}{2}$, 3+$\dfrac{1}{2}$ ) = (1, 2)
Area of a triangle = $\dfrac{1}{2 }× [x_{1} (y_{2} – y_{3} ) + x_{2} (y_{3} – y_{1} ) + x_{3} (y_{1} – y_{2} )]$
Area of $\triangle$ DEF = $\dfrac{1}{2}$ {1(2-1) + 1(1-0) + 0(0-2)} = $\dfrac{1}{2}$ (1+1) = 1
The area of $\triangle$ DEF is 1 square unit
Area of $\triangle$ ABC =$\dfrac{ 1}{2}$ [0(1-3) + 2{3-(-1)} + 0(-1-1)] = $\dfrac{1}{2}$ {8} = 4
The area of $\triangle$ABC is 4 square units
Therefore, the required ratio is 1:4.
Let the vertices of the quadrilateral be A (- 4, – 2), B ( – 3, – 5), C (3, – 2), and D (2, 3).
Join AC and divide the quadrilateral into two triangles.
We have two triangles, $\triangle$ ABC and $\triangle$ ACD.
Area of a triangle = $\dfrac{1}{2} × [x_{1} (y_{2} – y_{3} ) + x_{2} (y_{3} – y_{1} ) + x_{3} (y_{1} – y_{2} )]$
Area of $\triangle$ ABC = $\dfrac{1}{2}$ [(-4) {(-5) – (-2)} + (-3) {(-2) – (-2)} + 3 {(-2) – (-5)}]
= $\dfrac{1}{2}$ (12 + 0 + 9)
= $\dfrac{21}{2}$ square units
Area of $\triangle$ ACD = $\dfrac{1}{2}$ [(-4) {(-2) – (3)} + 3{(3) – (-2)} + 2 {(-2) – (-2)}]
= $\dfrac{1}{2}$ (20 + 15 + 0)
= $\dfrac{35}{2}$ square units
Area of quadrilateral ABCD = Area of $\triangle$ ABC + Area of $\triangle$ ACD
= ($\dfrac{21}{2}$ + $\dfrac{35}{2}$) square units = 28 square units
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 $\triangle$ ABC. Therefore, AD is the median in $\triangle$ ABC.
Coordinates of point D = Midpoint of BC = ($\dfrac{(3+5)}{2}$, $\dfrac{(-2+2)}{2}$) = (4, 0)
Formula to find the area of a triangle = $\dfrac{1}{2}× [x_{1} (y_{2} – y_{3} ) + x_{2} (y_{3} – y_{1} ) + x_{3} (y_{1} – y_{2} )]$
Now, the Area of $\triangle$ ABD = $\dfrac{1}{2}$ [(4) {(-2) – (0)} + 3{(0) – (-6)} + (4) {(-6) – (-2)}]
= $\dfrac{1}{2}$ (-8 + 18 – 16)
= -3 square units
However, the area cannot be negative. Therefore, the area of $\triangle$ ABD is 3 square units.
Area of $\triangle$ ACD = $\dfrac{1}{2}$ [(4) {0 – (2)} + 4{(2) – (-6)} + (5) {(-6) – (0)}]
= $\dfrac{1}{2}$ (-8 + 32 – 30) = -3 square units
However, the area cannot be negative. Therefore, the area of $\triangle$ ACD is 3 square units.
The area of both sides is the same. Thus, median AD has divided $\triangle$ ABC into two triangles of equal areas.