Question

The points A(2, 9), B(a, 5) and C(5, 5) are the vertices of a triangle ABC right angled at B. Find the values of a and hence the area of ΔABC.

Answer 1

Consider the figure.
Using distance formula,
$$\begin{gathered} \mathrm{AC}=\sqrt{{\left(2-5\right)}^2+{\left(9-5\right)}^2} \\ \mathrm{AC}=\sqrt{{\left(-3\right)}^2+{\left(4\right)}^2} \\ \mathrm{AC}=\sqrt{25} \\ {\mathrm{AC}}^2=25\mathrm{units} \end{gathered}$$
$$\begin{gathered} \mathrm{AB}=\sqrt{{\left(2-a\right)}^2+{\left(9-5\right)}^2} \\ \mathrm{AB}=\sqrt{\left(4+a^2-4a\right)+\left(16\right)} \\ \mathrm{AB}=\sqrt{4+a^2-4a+16}=\sqrt{a^2-4a+20} \\ {\mathrm{AB}}^2=(a^2-4a+20)\mathrm{units} \end{gathered}$$
$$\begin{gathered} \mathrm{BC}=\sqrt{{\left(5-a\right)}^2+{\left(5-5\right)}^2} \\ \mathrm{BC}=\sqrt{\left(25+a^2-10a\right)+0} \\ \mathrm{BC}=\sqrt{a^2-10a+25} \\ {\mathrm{BC}}^2=(a^2-10a+25)\mathrm{units} \end{gathered}$$
We are given that ABC is a right triangle right angled at B.
By Pythagoras theorem, we have;
$$\begin{gathered} {\mathrm{AC}}^2={\mathrm{AB}}^2+{\mathrm{BC}}^2 \\ \Rightarrow 25=(a^2-4a+20)+(a^2-10a+25) \\ \Rightarrow 25=2a^2-14a+45 \\ \Rightarrow 2a^2-14a+20=0 \\ \Rightarrow a^2-7a+10=0 \\ \Rightarrow a^2-5a-2a+10=0 \\ \Rightarrow a(a-5)-2(a-5)=0 \\ \Rightarrow (a-2)(a-5)=0 \\ \Rightarrow a=2ora=5 \end{gathered}$$
We cannot put a = 5 as it will make BC = 0. So, we ignore a = 5 and accept a = 2.
$$\begin{gathered} \Rightarrow {\mathrm{AB}}^2=4-8+20=16 \\ \Rightarrow \mathrm{AB}=4\mathrm{units} \\ \Rightarrow {\mathrm{BC}}^2=4+25-20=9 \\ \Rightarrow \mathrm{BC}=3\mathrm{units} \\ \mathrm{And},\mathrm{AC}=5\mathrm{units} \\ \mathrm{Area}=\frac{1}{2}\times \mathrm{base}\times \mathrm{height} \\ \mathrm{Area}=\frac{1}{2}\times 3\times 4=6\mathrm{square}\mathrm{units} \end{gathered}$$