Question

The vertices of a triangle are $(2,1),(5,2),$ and $(4,4)$. The lengths of the perpendicular from these vertices on the opposite sides are

• A. $\frac{7}{\sqrt{5}},\frac{7}{\sqrt{13}},\frac{7}{\sqrt{6}}$
• B. $\frac{7}{\sqrt{6}},\frac{7}{\sqrt{8}},\frac{7}{\sqrt{10}}$
• C. $\frac{7}{\sqrt{5}},\frac{7}{\sqrt{8}},\frac{7}{\sqrt{15}}$
• D. $\frac{7}{\sqrt{5}},\frac{7}{\sqrt{13}},\frac{7}{\sqrt{10}}$

Answer 1

The correct option is D

$\frac{7}{\sqrt{5}},\frac{7}{\sqrt{13}},\frac{7}{\sqrt{10}}$

Explanation for the correct option:

Let $A(2,1),B(5,2),$ and $C(4,4)$ are the vertices of the triangle.

Slope of $AB=\frac{(y_2–y_1)}{\left(x_2-x_1\right)}$

$$\begin{gathered} =\frac{(2–1)}{(5-2)} \\ =\frac{1}{3} \end{gathered}$$

Equation of line $AB\Rightarrow (y–y_1)=m(x–x_1)$

$$\begin{gathered} \Rightarrow (y–1)=\left(\frac{1}{3}\right)(x–2) \\ \Rightarrow 3y-3=x-2 \\ \Rightarrow x-3y+1=0 \end{gathered}$$

Length of the perpendicular from $C(4,4)$ to $AB=\frac{|a{x}_1+b{y}_1+c|}{\sqrt{a^2+b^2}}$

$$\begin{gathered} =\frac{|4-12+1|}{\sqrt{1^2+3^2}} \\ =\frac{7}{\sqrt{10}} \end{gathered}$$

Slope of $BC=\frac{4-2}{4-5}$

$=-2$

Equation of line $BC\Rightarrow (y–y_1)=m(x–x_1)$

$$\begin{gathered} \Rightarrow (y–2)=\left(-2\right)(x–5) \\ \Rightarrow 2x+y-12=0 \end{gathered}$$

Length of the perpendicular from $A(2,1)$ to $BC=\frac{|a{x}_1+b{y}_1+c|}{\sqrt{a^2+b^2}}$

$$\begin{gathered} =\frac{|4+1-12|}{\sqrt{2^2+1^2}} \\ =\frac{7}{\sqrt{5}} \end{gathered}$$

Slope of $AC=\frac{4-1}{4-2}$

$=\frac{3}{2}$

Equation of line $AC\Rightarrow (y–y_1)=m(x–x_1)$

$$\begin{gathered} \Rightarrow (y–1)=\left(\frac{3}{2}\right)(x–2) \\ \Rightarrow 2y-2=3x-6 \\ \Rightarrow 3x-2y-4=0 \end{gathered}$$

Length of the perpendicular from $B(5,2)$ to $AC=\frac{|a{x}_1+b{y}_1+c|}{\sqrt{a^2+b^2}}$

$$\begin{gathered} =\frac{|15-4-4|}{\sqrt{3^2+2^2}} \\ =\frac{7}{\sqrt{13}} \end{gathered}$$

The lengths of the perpendicular from the vertices on the opposite sides are $\frac{7}{\sqrt{5}},\frac{7}{\sqrt{13}},\frac{7}{\sqrt{10}}$.

Hence the correct option is $\mathrm{Option}\left(\mathrm{D}\right)$