Question

If $\Delta ABC$ is right-angled at B, where $A\left( 5,6,4 \right)$, $B\left( 4,4,1 \right)$, $C\left( 8,2,x \right)$, then find the value of $x$.

Answer 1

Hint: We first use the Pythagoras’ theorem that gives $bas{{e}^{2}}+heigh{{t}^{2}}=hypotenus{{e}^{2}}$. Then we get $A{{B}^{2}}+B{{C}^{2}}=A{{C}^{2}}$. We use the distance formula of $d=\sqrt{{{\left( x-a \right)}^{2}}+{{\left( y-b \right)}^{2}}+{{\left( z-c \right)}^{2}}}$. We find the length of the sides and use them in the formula of $A{{B}^{2}}+B{{C}^{2}}=A{{C}^{2}}$. We solve the linear equation to find the value of $x$.

Complete step by step solution:
$\Delta ABC$ is right-angled at B. Therefore, $\angle B={{90}^{\circ }}$ and the side AC is the hypotenuse.
From Pythagoras’ theorem we can say that $bas{{e}^{2}}+heigh{{t}^{2}}=hypotenus{{e}^{2}}$.
Here the rest of the sides AB and BC will be considered as height and base.
We get $A{{B}^{2}}+B{{C}^{2}}=A{{C}^{2}}$.

We have been given the coordinates of the vertices.
For two points with coordinates $P\left( x,y,z \right)$, $Q\left( a,b,c \right)$, the distance between them will be
$d=\sqrt{{{\left( x-a \right)}^{2}}+{{\left( y-b \right)}^{2}}+{{\left( z-c \right)}^{2}}}$.
We find the length of the sides AB, BC, CA.
So, $AB=\sqrt{{{\left( 5-4 \right)}^{2}}+{{\left( 6-4 \right)}^{2}}+{{\left( 4-1 \right)}^{2}}}=\sqrt{14}$
\[AC=\sqrt{{{\left( 5-8 \right)}^{2}}+{{\left( 6-2 \right)}^{2}}+{{\left( 4-x \right)}^{2}}}=\sqrt{{{\left( x-4 \right)}^{2}}+25}\]
\[BC=\sqrt{{{\left( 4-8 \right)}^{2}}+{{\left( 4-2 \right)}^{2}}+{{\left( 1-x \right)}^{2}}}=\sqrt{{{\left( x-1 \right)}^{2}}+20}\]
The equation $A{{B}^{2}}+B{{C}^{2}}=A{{C}^{2}}$ gives $14+{{\left( x-1 \right)}^{2}}+20={{\left( x-4 \right)}^{2}}+25$.
We simplify the quadratic to get
$\begin{align}
  & 14+{{\left( x-1 \right)}^{2}}+20={{\left( x-4 \right)}^{2}}+25 \\
 & \Rightarrow -2x+35=-8x+41 \\
 & \Rightarrow 8x-2x=41-35 \\
 & \Rightarrow 6x=6 \\
\end{align}$
We now divide both sides with 6 to get $x=\dfrac{6}{6}=1$.
Therefore, the value of $x$ is 1.

Note: We need to be careful about taking the image as 2-D as it is situated in a 3-D plane. The length of the sides will be considered in that way also. The hypotenuse also has to be greater than any other sides and that can be considered as a condition also.