Find the area of triangle ABC whose vertices are A(-5,7) ,B(-4,-5) and C (4,5).
Last updated date: 24th Mar 2023
•
Total views: 306.6k
•
Views today: 8.83k
Answer
306.6k+ views
Hint: To calculate the area of triangle, we use the formula,
Area = \[\dfrac{1}{2}\left[ {{x}_{1}}~\left( {{y}_{2-}}~{{y}_{3}}~ \right)+{{x}_{2}}~\left( {{y}_{3}}-{{y}_{1}}~ \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right]\]-- (1)
Where, $({{x}_{1}},{{y}_{1}}),({{x}_{2}},{{y}_{2}}),({{x}_{3}},{{y}_{3}})$are the coordinates of A, B and C.
Complete step-by-step answer:
We can calculate the area by using the area formula. One should however remember that since area is a positive quantity, we take the absolute value of the area we get from formula (1).
Thus, we get,
Area = \[\dfrac{1}{2}\left[ {{x}_{1}}~\left( {{y}_{2-}}~{{y}_{3}}~ \right)+{{x}_{2}}~\left( {{y}_{3}}-{{y}_{1}}~ \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right]\]
Now, we know the coordinates of A, B and C which are (-5, 7), (-4, -5) and (4, 5) respectively. These are put in place of $({{x}_{1}},{{y}_{1}}),({{x}_{2}},{{y}_{2}}),({{x}_{3}},{{y}_{3}})$. Here we have taken the order of$({{x}_{1}},{{y}_{1}}),({{x}_{2}},{{y}_{2}}),({{x}_{3}},{{y}_{3}})$ as A, B and C. However, one can also take any other order (say B, C and A). We should however be consistent while inserting the values. Now, we start solving,
Area = $\dfrac{1}{2}$[-5(-5-5) + {-4(5-7)} + 4{7-(-5)}]
Area = $\dfrac{1}{2}$[50+8+48]
Area = 53 square units.
Hence, the area of triangle ABC is 53 square units.
Note: While calculating the area of the triangle when the Cartesian coordinates are given, one can also proceed by first plotting the triangle on X-Y graph. This can help in identifying if the triangle is an equilateral triangle, isosceles triangle or right triangle. If we can identify that the triangle is one of them, we can easily calculate the area of the triangle, by using the respective formulas for these special types of triangles. This greatly reduces the time taken in calculating the area of the triangle. In case, the triangle is none of the above types of triangles, we can always use the normal formula in Cartesian coordinates to calculate the area.
Area = \[\dfrac{1}{2}\left[ {{x}_{1}}~\left( {{y}_{2-}}~{{y}_{3}}~ \right)+{{x}_{2}}~\left( {{y}_{3}}-{{y}_{1}}~ \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right]\]-- (1)
Where, $({{x}_{1}},{{y}_{1}}),({{x}_{2}},{{y}_{2}}),({{x}_{3}},{{y}_{3}})$are the coordinates of A, B and C.
Complete step-by-step answer:
We can calculate the area by using the area formula. One should however remember that since area is a positive quantity, we take the absolute value of the area we get from formula (1).
Thus, we get,
Area = \[\dfrac{1}{2}\left[ {{x}_{1}}~\left( {{y}_{2-}}~{{y}_{3}}~ \right)+{{x}_{2}}~\left( {{y}_{3}}-{{y}_{1}}~ \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right]\]
Now, we know the coordinates of A, B and C which are (-5, 7), (-4, -5) and (4, 5) respectively. These are put in place of $({{x}_{1}},{{y}_{1}}),({{x}_{2}},{{y}_{2}}),({{x}_{3}},{{y}_{3}})$. Here we have taken the order of$({{x}_{1}},{{y}_{1}}),({{x}_{2}},{{y}_{2}}),({{x}_{3}},{{y}_{3}})$ as A, B and C. However, one can also take any other order (say B, C and A). We should however be consistent while inserting the values. Now, we start solving,
Area = $\dfrac{1}{2}$[-5(-5-5) + {-4(5-7)} + 4{7-(-5)}]
Area = $\dfrac{1}{2}$[50+8+48]
Area = 53 square units.
Hence, the area of triangle ABC is 53 square units.
Note: While calculating the area of the triangle when the Cartesian coordinates are given, one can also proceed by first plotting the triangle on X-Y graph. This can help in identifying if the triangle is an equilateral triangle, isosceles triangle or right triangle. If we can identify that the triangle is one of them, we can easily calculate the area of the triangle, by using the respective formulas for these special types of triangles. This greatly reduces the time taken in calculating the area of the triangle. In case, the triangle is none of the above types of triangles, we can always use the normal formula in Cartesian coordinates to calculate the area.
Recently Updated Pages
If ab and c are unit vectors then left ab2 right+bc2+ca2 class 12 maths JEE_Main

A rod AB of length 4 units moves horizontally when class 11 maths JEE_Main

Evaluate the value of intlimits0pi cos 3xdx A 0 B 1 class 12 maths JEE_Main

Which of the following is correct 1 nleft S cup T right class 10 maths JEE_Main

What is the area of the triangle with vertices Aleft class 11 maths JEE_Main

The coordinates of the points A and B are a0 and a0 class 11 maths JEE_Main

Trending doubts
Write an application to the principal requesting five class 10 english CBSE

Tropic of Cancer passes through how many states? Name them.

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE

What is per capita income

Change the following sentences into negative and interrogative class 10 english CBSE

A Short Paragraph on our Country India
