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Question: Find the area of triangle ABC, coordinates of whose vertices are \(A\left( 4,1 \right),B\left( 6,6 \...

Find the area of triangle ABC, coordinates of whose vertices are A(4,1),B(6,6)A\left( 4,1 \right),B\left( 6,6 \right) and C(8,4)C\left( 8,4 \right).

Explanation

Solution

Assume the coordinates of vertices A(4,1),B(6,6)A\left( 4,1 \right),B\left( 6,6 \right) and C(8,4)C\left( 8,4 \right) as (x1,y1),(x2,y2)\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right) and (x3,y3)\left( {{x}_{3}},{{y}_{3}} \right) respectively. Apply the formula for area of triangle in coordinate form given as : Area=12[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]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] . If the area turns out to be positive then that will be the required answer but if it turns out to be negative then take modulus so that it becomes positive.

Complete step-by-step solution:
Here, we have been provided with a triangle ABC whose vertices are given as A(4,1),B(6,6)A\left( 4,1 \right),B\left( 6,6 \right) and C(8,4)C\left( 8,4 \right).

Now, assuming the coordinates of A, B and C as (x1,y1),(x2,y2)\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right) and (x3,y3)\left( {{x}_{3}},{{y}_{3}} \right), we get,
A(4,1)=(x1,y1) B(6,6)=(x2,y2) C(8,4)=(x3,y3) \begin{aligned} & A\left( 4,1 \right)=\left( {{x}_{1}},{{y}_{1}} \right) \\\ & B\left( 6,6 \right)=\left( {{x}_{2}},{{y}_{2}} \right) \\\ & C\left( 8,4 \right)=\left( {{x}_{3}},{{y}_{3}} \right) \\\ \end{aligned}
Now, we know that area of triangle with coordinates (x1,y1),(x2,y2)\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right) and (x3,y3)\left( {{x}_{3}},{{y}_{3}} \right) is given by the expression :
Area=12[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]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, substituting the values of x1,x2,x3,y1,y2{{x}_{1}},{{x}_{2}},{{x}_{3}},{{y}_{1}},{{y}_{2}} and y3{{y}_{3}}in the above expression for area, we get,
Ar.(ΔABC)=12[4(6−4)+6(4−1)+8(1−6)] ⇒Ar.(ΔABC)=12[4×2+6×3+8×(−5)] ⇒Ar.(ΔABC)=12[8+18−40] ⇒Ar.(ΔABC)=−142 ⇒Ar.(ΔABC)=−7 square units \begin{aligned} & Ar.\left( \Delta ABC \right)=\dfrac{1}{2}\left[ 4\left( 6-4 \right)+6\left( 4-1 \right)+8\left( 1-6 \right) \right] \\\ & \Rightarrow Ar.\left( \Delta ABC \right)=\dfrac{1}{2}\left[ 4\times 2+6\times 3+8\times \left( -5 \right) \right] \\\ & \Rightarrow Ar.\left( \Delta ABC \right)=\dfrac{1}{2}\left[ 8+18-40 \right] \\\ & \Rightarrow Ar.\left( \Delta ABC \right)=\dfrac{-14}{2} \\\ & \Rightarrow Ar.\left( \Delta ABC \right)=-7\text{ square units} \\\ \end{aligned}
As we can see that the area of the ΔABC\Delta ABC turns out to be a negative value and we know that area cannot be negative. So, we must take the modulus on both sides to get the value of the area positive. Therefore, taking modulus, we get,
∴Ar.(ΔABC)=∣−7∣=7 square units\therefore Ar.\left( \Delta ABC \right)=\left| -7 \right|=7\text{ square units}
Hence, the area of the ΔABC\Delta ABC is 7 square units.

Note: One may note that the formula for area of triangle in simplified form, can also be given as : Area=12[(x1y2−y1x2)+(x2y3−y2x3)+(x3y1−y3x1)]Area=\dfrac{1}{2}\left[ \left( {{x}_{1}}{{y}_{2}}-{{y}_{1}}{{x}_{2}} \right)+\left( {{x}_{2}}{{y}_{3}}-{{y}_{2}}{{x}_{3}} \right)+\left( {{x}_{3}}{{y}_{1}}-{{y}_{3}}{{x}_{1}} \right) \right]. Always remember that if by substituting the values of given coordinates in the formula for area, the area turns out to be negative then we have to take modulus to make it positive. Actually we can choose any of the vertices as (x1,y1),(x2,y2)\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right) and (x3,y3)\left( {{x}_{3}},{{y}_{3}} \right). It will not change the answer.