Question

A(–3, 0), B(4,–1) and C(5, 2) are vertices of ΔABC. The length of altitude through A is __________.

Answer 1

In ΔABC, Let D be such that AD defines altitude of ΔABC.
A = (–3, 0), B = (4, –1) and C = (5, 2)

$\mathrm{Area}\mathrm{of}∆ABC=\frac{1}{2}\left|\begin{array}{ccc}-3& 0& 1\\ 4& -1& 1\\ 5& 2& 1\end{array}\right|$ applying R₂ → R₂ – R₁ and R₃ → R₃ – R_{1
$$\begin{gathered} \mathrm{i}.\mathrm{e}\frac{1}{2}\left|\begin{array}{ccc}-3& 0& 1\\ 7& -1& 0\\ 8& 2& 0\end{array}\right| \\ \mathrm{i}.\mathrm{e}\frac{1}{2}\left(14+8\right)=\frac{22}{2}=11 \end{gathered}$$}
Since area of ΔABC = $\frac{1}{2}$ × Base × Altitude

$$\begin{gathered} \mathrm{Here}\mathrm{Base}=BC=\sqrt{{\left(5-4\right)}^2+{\left(2+1\right)}^2} \\ =\sqrt{1+9}=\sqrt{10} \\ \Rightarrow 11=\frac{1}{2}\times \sqrt{10}\times AD \\ \mathrm{i}.\mathrm{e}\frac{22}{\sqrt{10}}\mathrm{is}\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{altitude}AD. \end{gathered}$$