Coordinate Geometry Chapter 3 Area Of Triangular Region
Area of formula:
1. The area of ΔABC where vertices are A (x1, y1), B (x2, y2) and C (x3, y3) respectively is \(\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|\) sq. unit
or, = \(\frac{1}{2}\left|y_1\left(x_2-x_3\right)+y_2\left(x_3-x_1\right)+y_3\left(x_1-x_2\right)\right|\) sq. unit [Area is always taken positive]
2. The points (x1, y1), (x2, y2), (y3, y3) are collinear if x1 (y2 – y3)+ x2 (y3 – y1) + x3 (y1 – y2) = 0
or, y1 (x2 – x3) + y2 (x3 – x1) + y3(x1 – x2) = 0
Read and Learn More WBBSE Solutions For Class 9 Maths
Area of the triangular region is
= \(\frac{1}{2}=\left|x_1 y_2+x_2 y_3+x_3 y_1-\left(y_1 x_2+y_2 x_3+y_3 x_1\right)\right|\) sq. unit
Similarly, area of quadrilateral region is
= \(\frac{1}{2}\left|\left(x_1 y_2+x_2 y_3+x_3 y_4+x_4 y_1\right)-\left(y_1 x_2+y_2 x_3+y_3 x_4+y_4 x_1\right)\right|\) sq. unit
Coordinate Geometry Chapter 3 Area Of Triangular Region Fill In The Blanks
Example 1. Area of the triangle having vertices (2, 2), (4, 2), (1, 3) is _______ sq. unit.
Solution: 11
Example 2. (0, -2), (2, 4), (-1, -5) are ______ points.
Solution: Collinear.
Example 3. If (2, -1), (k, -1), (1, -1) are collinear points then k = ________
Solution: Any real value.
Example 4. If the points (1, 2), (2, 4), and (6, 6) are collinear then t =
Solution: 3
Example 5. If the vertices of a triangle are (-1, 0), (0, 0), (0, 1) then its area is ______ sq. unit
Solution: \(\frac{1}{2}\)
Example 6. If the three points (0, 0), (2, -3), (x, y) are collinear then x = ______, y = _______
Solution: 4, -6.
Coordinate Geometry Chapter 3 Area Of Triangular Region True Or False
Example 1. Area of the triangle having vertices (-3, 1), (– 2, 2), and (3, 2) is 9\(\frac{1}{2}\) sq. unit.
Solution: The statement is true.
Example 2. If the vertices of a triangle are (a, b + c), (a, b – c), and (-a, c) then area is \(\frac{1}{2}\)/ab/sq.unit
Solution: The statement is false.
Example 3. (1, 4), (-1, 2), (-4, -1) are collinear points.
Solution: The statement is true.
Example 4. (a, b + c), (b, c + a), (c, a + b) are not collinear points.
Solution: The statement is false.
Example 5. Value of k for which (k, k), (-1, 5), (-7, 8) be collinear is 3.
Solution: The statement is true.
Example 6. If the points (a, c), (0, b), (\(\frac{1}{2}\), \(\frac{1}{2}\)) are collinear points then \(\frac{1}{a}+\frac{1}{b}=2\)
Solution: The statement is false.
Coordinate Geometry Chapter 3 Area Of Triangular Region Short Answer Type Questions
Example 1. If the three points (a, 0), (0, b), (1, 1) are collinear then show that \(\frac{1}{a}+\frac{1}{b}=1\)
Solution:
= \(\frac{1}{2}|a b+0+0-b-a|=0 \quad \text { or, } a+b=a b\)
∴ \(\frac{1}{a}+\frac{1}{h}=1\)
Example 2. Find the area of the triangular area region formed by the three points (1, 4), (-1, 2), and (-4, 1).
Solution:
⇒ Area = \(\frac{1}{2}|2-1-16+4+8-1| \text { sq.unit }=\frac{1}{2}|14-18| \text { sq.. unit }=2\)
The area of the triangular area region = \(\frac{1}{2}|2-1-16+4+8-1| \text { sq.unit }=\frac{1}{2}|14-18| \text { sq.. unit }=2\)
Example 3. If the points (0, -4), (-1, y), and (3, 2) are on the same straight line find y.
Solution: \(\frac{1}{2}|0(y-2)+(-1)\{2-(-4)\}+3(-4-4)|=0\)
⇒ or, -6 – 12 – 3y = 0, y = 6
Example 4. If the points A (3a, k), B (0, 3b), C (a, 2b) are collinear find k.
Solution: 3a (3b – 2b) + 0 (2b – k) + a (k – 3b) = 0
or, 3ab + ak – 3ab 0 or, ak = 0
∴ k = 0
Example 5. Show that (a, b + c), (b, c + a), and (c, a + b) are collinear.
Solution: \(\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|\)
= \(\frac{1}{2}[a(c+a-a-b)+b(a+b-b-c)+c(b+c-c-a)]\)
= \(\frac{1}{2}(a c-a b+a b-b c+b c-a c)=0\)
⇒ Here x1 = a, y1 = b + c
⇒ x2 = b, y2 = c + a.
⇒ x3 = c, y3 = a + b
Example 6. If A (x, 4), B (- 5, 7), C (-4, 5) are collinear, then show that 2x+y+3=0
Solution: \(\frac{1}{2}\)[x(7-5) + (-5)(5 − y) + (−4)(y −7)] = 0
⇒ or, \(\frac{1}{2}\)(2x+y+3)=0
∴ 2x + y + 3 = 0