Solveeit Logo
Login

Question

Mathematics Question on Determinants

A triangle has its three sides equal to p, q and r. If the coordinates of its vertices are A(x1,y1)A\left(x_{1}, y_{1}\right), B(x2,y2)B\left(x_{2}, y_{2}\right) and C(x3,y3)C\left(x_{3}, y_{3}\right), then ∣x1y12 x2y22 x3y32∣2\left|\begin{matrix}x_{1}&y_{1}&2\\\ x_{2}&y_{2}&2\\\ x_{3}&y_{3}&2\end{matrix}\right|^{2} is equal to

A

(p+q+r)\left(p+q+r\right)

B

(p+q−r)2\left(p+q-r\right)^{2}

C

(p+q+r)(q+r−p)(r+p−q)(p+q−r)\left(p+q+r\right) \left(q+r-p\right)\left(r+p-q\right)\left(p+q-r\right)

D

(p−q+r)(q+r−p)(p+q−r)\left(p-q+r\right) \left(q+r-p\right)\left(p+q-r\right)

Answer

(p+q+r)(q+r−p)(r+p−q)(p+q−r)\left(p+q+r\right) \left(q+r-p\right)\left(r+p-q\right)\left(p+q-r\right)

Explanation

Solution

Let Δ\Delta be the area of triangle ABCABC. Then, Δ=12∣x1y11 x2y21 x3y31∣\Delta=\frac{1}{2}\left|\begin{matrix}x_{1}&y_{1}&1\\\ x_{2}&y_{2}&1\\\ x_{3}&y_{3}&1\end{matrix}\right| ⇒2Δ=∣x1y11 x2y21 x3y31∣⇒4Δ=2∣x1y11 x2y21 x3y31∣\Rightarrow\quad2\Delta=\left|\begin{matrix}x_{1}&y_{1}&1\\\ x_{2}&y_{2}&1\\\ x_{3}&y_{3}&1\end{matrix}\right| \Rightarrow 4\Delta=2\left|\begin{matrix}x_{1}&y_{1}&1\\\ x_{2}&y_{2}&1\\\ x_{3}&y_{3}&1\end{matrix}\right| ⇒4Δ=∣x1y12 x2y22 x3y32∣⇒16Δ2=∣x1y12 x2y22 x3y32∣2…(i)\Rightarrow\quad4\Delta=\left|\begin{matrix}x_{1}&y_{1}&2\\\ x_{2}&y_{2}&2\\\ x_{3}&y_{3}&2\end{matrix}\right|\Rightarrow 16\Delta^{2}=\left|\begin{matrix}x_{1}&y_{1}&2\\\ x_{2}&y_{2}&2\\\ x_{3}&y_{3}&2\end{matrix}\right|^{2} \quad\ldots\left(i\right) We also know that the area of triangle ABC is given by Δ=s(s−p)(s−q)(s−r),wheres=12(p+q+r)\Delta=\sqrt{s\left(s-p\right)\left(s-q\right)\left(s-r\right)}, where s =\frac{1}{2}\left(p+q+r\right) But, s=12(p+q+r)⇒s−p=12(p+q+r)−p=12(q+r−p)s=\frac{1}{2} \left(p+q+r\right)\Rightarrow s-p=\frac{1}{2}\left(p+q+r\right)-p=\frac{1}{2}\left(q+r-p\right) Similarly, s−q=12(r+p−q)ands−r=12(p+q−r)s-q=\frac{1}{2}\left(r+p-q\right)and s - r =\frac{1}{2}\left(p+q-r\right) ∴Δ2=12(p+q+r)⋅12(q+r−p)⋅12(r+p−q)⋅12(p+q−r)\therefore \Delta^{2}=\frac{1}{2}\left(p+q+r\right)\cdot\frac{1}{2}\left(q+r-p\right)\cdot\frac{1}{2} \left(r+p-q\right)\cdot\frac{1}{2}\left(p+q-r\right) ⇒16Δ2=(p+q+r)(q+r−p)(r+p−q)(p+q−r)…(ii)\Rightarrow\quad16\Delta^{2}=\left(p+q+r\right)\left(q+r-p\right)\left(r+p-q\right)\left(p+q-r\right)\quad\ldots\left(ii\right) From (i)and(ii)\left(i\right) and \left(ii\right), we get ∣x1y12 x2y22 x3y32∣2=(p+q+r)(q+r−p)(r+p−q)(p+q−r)\left|\begin{matrix}x_{1}&y_{1}&2\\\ x_{2}&y_{2}&2\\\ x_{3}&y_{3}&2\end{matrix}\right|^{2}=\left(p + q + r\right)\left(q+r-p\right)\left(r+p-q\right)\left(p+q-r\right)