Question

Show that the points A (0, 6), B (2, 1) and C (7, 3) are three corners of a square ABCD.
Find (i) the slope of the diagonal BD and,
(ii) the coordinates of the fourth vertex D.

Answer 1

Hint: We will first start by using the distance formula to find the distance between all the three pairs of points then we will use the Pythagoras theorem to show that the triangle is right angle also we will use the fact that if two sides of the triangle are equal then the triangle is an isosceles triangle also we will use the fact that the diagonal of a square made an right angle isosceles triangle with the sides of the square so we have the three vertices as the three vertices of the square then we will use the formula for finding the slope of the diagonal of the square and use the midpoint formula to find the coordinates of the fourth vertex.

Complete step by step answer:

Now we have the three points as A (0, 6), B (2, 1) and C (7, 3) now we know that according to the distance formula the distance between two points$\left( {{x}_{1}},{{x}_{2}} \right)\ and\ \left( {{y}_{1}},{{y}_{2}} \right)$is $\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+\ {{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$ therefore we have the distances as

$\begin{align}

  & AB=\sqrt{{{\left( 0-2 \right)}^{2}}+{{\left( 6-1 \right)}^{2}}}=\sqrt{4+25}=\sqrt{29} \\

 & BC=\sqrt{{{\left( 2-7 \right)}^{2}}+{{\left( 1-3 \right)}^{2}}}=\sqrt{{{\left( -5 \right)}^{2}}+{{\left( -2 \right)}^{2}}}=\sqrt{25+4}=\sqrt{29} \\

 & AC=\sqrt{{{\left( 0-7 \right)}^{2}}+{{\left( 6-3 \right)}^{2}}}=\sqrt{{{\left( -7 \right)}^{2}}+{{\left( 3 \right)}^{2}}}=\sqrt{49+9}=\sqrt{58} \\

\end{align}$

Now we can see that $A{{B}^{2}}+B{{C}^{2}}=A{{C}^{2}}$also AB = BC. Therefore, by converse of Pythagoras theorem we can say that the triangle ABC is a right angle isosceles triangle and since we know that the triangle formed by the diagonals of a square with its sides is an isosceles right angle triangle. Therefore, we have that the vertices A, B and C are the vertices of the a square.

Now, we have been given three points of the vertices of square ABCD as,

We let the fourth coordinate D be (x, y)

Now, we know that the diagonals of squares are perpendicular to each other. Also, we know that if two lines with slope ${{m}_{1}}\ and\ {{m}_{2}}$ are perpendicular then ${{m}_{1}}{{m}_{2}}=-1$.

Now, we know that the slope of line joining $\left( {{x}_{1}},{{y}_{1}} \right)\ and\ \left( {{x}_{2}},{{y}_{2}} \right)$ is $\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$.

So, we have slope of $AB={{m}_{1}}$

$\begin{align}

  & =\dfrac{6-3}{0-7} \\

 & =\dfrac{3}{-7} \\

 & =\dfrac{-3}{7} \\

\end{align}$

Now, we have the slope of $BC={{m}_{2}}=\dfrac{y-1}{x-2}.........\left( 1 \right)$

Also, we know that for perpendicular lines,

$\begin{align}

  & {{m}_{1}}\times {{m}_{2}}=-1 \\

 & \dfrac{-3}{7}\times {{m}_{2}}=-1 \\

 & {{m}_{2}}=\dfrac{7}{3}..........\left( 2 \right) \\

\end{align}$

Now, we know that the midpoint of diagonal AC = midpoint of diagonal BD

$\left( \dfrac{0+7}{2},\dfrac{6+3}{2} \right)=\left( \dfrac{x+2}{2},\dfrac{y+1}{2} \right)$

Now, on comparing we have,

$\begin{align}

  & \dfrac{7}{2}=\dfrac{x+2}{2} \\

 & x=5 \\

 & \dfrac{6+3}{2}=\dfrac{y+1}{2} \\

 & y=8 \\

\end{align}$

So, the coordinate of D is (5, 8) and the slope of BD is $\dfrac{7}{3}$.

Note: To solve these types of questions it is important to note that we have used the fact that the diagonals are perpendicular and midpoint of diagonal AC = midpoint of diagonal BD also it is important to note that how we have first proved the triangle formed by the three vertices as right angle isosceles triangle and we know that the triangle formed by the diagonal of the square with its side is also an right angle isosceles triangle hence we have shown that the three vertices are that of a square.