Question

Line through the points \[\left( -2,6 \right)\] and \[\left( 4,8 \right)\] is perpendicular to the line through the points
\[\left( 8,12 \right)\] and \[\left( x,24 \right)\] then find the value of \['x'\]

Answer 1

Hint: We solve this problem first by finding the slopes of two lines. The formula of slope of line having two points as \[A\left( {{x}_{1}},{{y}_{1}} \right),B\left( {{x}_{2}},{{y}_{2}} \right)\] is given as
\[m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\]
Then we use the condition that if the lines having slopes \[{{m}_{1}},{{m}_{2}}\] are perpendicular then
\[\Rightarrow {{m}_{1}}\times {{m}_{2}}=-1\]
By using this condition we find the value of \['x'\]

Complete step by step solution:
We are given that there are two lines.
We are given that the two points that lie on first line are \[\left( -2,6 \right)\] and \[\left( 4,8 \right)\]
Let us assume that the slope of first line as \[{{m}_{1}}\]
We know that the formula of slope of line having two points as \[A\left( {{x}_{1}},{{y}_{1}} \right),B\left( {{x}_{2}},{{y}_{2}} \right)\] is given as
\[m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\]
By using the above formula to first line we get
\[\begin{align}
  & \Rightarrow {{m}_{1}}=\dfrac{8-6}{4-\left( -2 \right)} \\
 & \Rightarrow {{m}_{1}}=\dfrac{2}{6} \\
 & \Rightarrow {{m}_{1}}=\dfrac{1}{3} \\
\end{align}\]
We are given that the two points that lie on second line are \[\left( 8,12 \right)\] and \[\left( x,24 \right)\]
Let us assume that the slope of first line as \[{{m}_{2}}\]
We know that the formula of slope of line having two points as \[A\left( {{x}_{1}},{{y}_{1}} \right),B\left( {{x}_{2}},{{y}_{2}} \right)\] is given as
\[m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\]
By using the above formula to second line we get
\[\begin{align}
  & \Rightarrow {{m}_{2}}=\dfrac{24-12}{x-8} \\
 & \Rightarrow {{m}_{2}}=\dfrac{12}{x-8} \\
\end{align}\]
We are given that the two lines are perpendicular to each other.
We know that if the lines having slopes \[{{m}_{1}},{{m}_{2}}\] are perpendicular then
\[\Rightarrow {{m}_{1}}\times {{m}_{2}}=-1\]
By using the above theorem to given two lines we get
\[\begin{align}
  & \Rightarrow \left( \dfrac{1}{3} \right)\left( \dfrac{12}{x-8} \right)=-1 \\
 & \Rightarrow \dfrac{12}{3x-24}=-1 \\
\end{align}\]
Now, by cross multiplying the terms in above equation we get
\[\begin{align}
  & \Rightarrow 3x-24=-12 \\
 & \Rightarrow 3x=12 \\
 & \Rightarrow x=4 \\
\end{align}\]
Therefore, the value of \['x'\] is given as
\[\therefore x=4\]

Note: Students may make mistakes in the slope formula.
We have the formula of slope of line having two points as \[A\left( {{x}_{1}},{{y}_{1}} \right),B\left( {{x}_{2}},{{y}_{2}} \right)\] is given as
\[m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\]
But due to confusion or in a hurry students may take the formula as
\[m=\dfrac{{{y}_{1}}-{{y}_{2}}}{{{x}_{2}}-{{x}_{1}}}\]
Or else they may take the formula as
\[m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{1}}-{{x}_{2}}}\]
This gives the wrong answer.
So, the formula of the slope has to be noted correctly.