Question

The length of altitude through A of $\Delta ABC$ where A (-3, 0), B (4, -1), C (5, 2) is:
a). $\dfrac{2}{\sqrt{10}}$
b). $\dfrac{4}{\sqrt{10}}$
c). $\dfrac{11}{\sqrt{10}}$
d). $\dfrac{22}{\sqrt{10}}$

Answer 1

Hint: We are going to find the area of $\Delta ABC$ using the determinant form and then equate this area of the triangle to half of the product of base and altitude. And the length of the base is calculated by distance formula between B and C.

Complete step-by-step solution -
In the below diagram, we have shown $\Delta ABC$ with coordinates A (-3, 0), B (4, -1) and C (5, 2) along with the altitude AD.

The area of $\Delta ABC$ with $ A (x_1, y_1), B (x_2, y_2), C (x_3, y_3) $ in the determinant form is given below:

$\Delta =\dfrac{1}{2}\begin{vmatrix}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1\end{vmatrix}$

Now, substituting the values of A (-3, 0), B (4, -1), C (5, 2) in the above determinant we get:
$\Delta =\dfrac{1}{2}\begin{vmatrix}
-3 & 0 & 1 \\
4 & -1 & 1 \\
5 & 2 & 1\end{vmatrix}$

Applying the row transformations in the above determinant $ R_2 $ → $ R_2 – R_1 $ and $ R_3 $ → $ R_3 – R_1 $ (where R is the symbol for row) we get:
$\Delta =\dfrac{1}{2}\begin{vmatrix}
-3 & 0 & 1 \\
7 & -1 & 1 \\
8 & 2 & 1\end{vmatrix}$

Now, expanding the determinant along third column we get:

$\begin{align}
  & \Delta =\dfrac{1}{2}\left| 14+8 \right| \\
 & \Rightarrow \Delta =\dfrac{1}{2}\left| 22 \right| \\
\end{align}$
We also know that:
 $\text{Area of }\Delta \text{ABC = }\dfrac{1}{2}\left( \text{base} \right)\left( \text{altitude} \right)$
Now, the length of base BC is calculated by distance formula.
Distance formula between $ B (x_1, y_1) $ and $ C (x_2, y_2) $ is given below:
BC = $\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$
Substituting the values of B (4, -1) and C (5, 2) in the above formula will lead to:
$\begin{align}
  & BC=\sqrt{{{\left( 5-4 \right)}^{2}}+{{\left( 2+1 \right)}^{2}}} \\
 & \Rightarrow BC=\sqrt{{{\left( 1 \right)}^{2}}+{{\left( 3 \right)}^{2}}} \\
 & \Rightarrow BC=\sqrt{10} \\
\end{align}$
Now, equating the area of $\Delta ABC$ in determinant form with the area of $\Delta ABC$ in the form of altitude and base we get,
$\dfrac{1}{2}\left| 22 \right|=\dfrac{1}{2}\left( \sqrt{10} \right)\left( \text{altitude} \right)$
One half will be cancelled from both the sides so altitude is equal to$\dfrac{22}{\sqrt{10}}$.
Hence, the correct option is (d).

Note: The area of a triangle is always positive so if the value of determinant in the area calculation is coming negative then remove the negative sign and just take the positive value of the determinant.