Answer
Verified
390.3k+ views
Hint: We start solving the problem by drawing the given information to get a better view. We then find the equation of the sides of the triangle where the line $x=c$ cuts the sides. We then find the area of the given triangle using the given vertices. We then find the points at which the line $x=c$ cuts the sides of the triangle. We then find the area of the region lying between one of the vertices and the intersect points to equate with half the area of the triangle to get the required value of c.
Complete step-by-step answer:
According to the problem, we have a triangle with vertices $\left( 0,0 \right)$, $\left( 1,1 \right)$, $\left( 9,1 \right)$ and a line $x=c$ cuts this triangle into two regions. We need to find the value of c, if the areas of two regions are the same.
We know that the equation of y-axis is $x=0$. So, the line $x=c$ is parallel to x-axis.
Let us draw the given information to get a better view.
Complete step-by-step answer:
According to the problem, we have a triangle with vertices $\left( 0,0 \right)$, $\left( 1,1 \right)$, $\left( 9,1 \right)$ and a line $x=c$ cuts this triangle into two regions. We need to find the value of c, if the areas of two regions are the same.
We know that the equation of y-axis is $x=0$. So, the line $x=c$ is parallel to x-axis.
Let us draw the given information to get a better view.
Let us find the equation of the side’s AC and BC.
We know that the equation of the line passing through two points $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ is $\left( y-{{y}_{1}} \right)=\dfrac{\left( {{y}_{2}}-{{y}_{1}} \right)}{\left( {{x}_{2}}-{{x}_{1}} \right)}\times \left( x-{{x}_{1}} \right)$.
We have the line segment AC passing through $A\left( 0,0 \right)$ and $C\left( 9,1 \right)$.
So, the equation of side AC is $\left( y-0 \right)=\dfrac{\left( 1-0 \right)}{\left( 9-0 \right)}\times \left( x-0 \right)$.
$\Rightarrow y=\dfrac{1}{9}\times x$.
$\Rightarrow y=\dfrac{x}{9}$ -(1).
We have the line segment BC passing through $B\left( 1,1 \right)$ and $C\left( 9,1 \right)$.
So, the equation of side BC is $\left( y-1 \right)=\dfrac{\left( 1-1 \right)}{\left( 9-1 \right)}\times \left( x-1 \right)$.
$\Rightarrow y-1=\dfrac{0}{8}\times \left( x-1 \right)$.
$\Rightarrow y=1$ -(2).
Now, let us find the area of the triangle ABC. Let us assume it as ${{A}_{1}}$.
We know that the area of the triangle with vertices $\left( {{x}_{1}},{{y}_{1}} \right)$, $\left( {{x}_{2}},{{y}_{2}} \right)$ and $\left( {{x}_{3}},{{y}_{3}} \right)$ is $\dfrac{1}{2}\left| \begin{matrix}
{{x}_{2}}-{{x}_{1}} & {{y}_{2}}-{{y}_{1}} \\
{{x}_{3}}-{{x}_{1}} & {{y}_{3}}-{{y}_{1}} \\
\end{matrix} \right|$.
$\Rightarrow {{A}_{1}}=\dfrac{1}{2}\left| \begin{matrix}
1-0 & 1-0 \\
9-0 & 1-0 \\
\end{matrix} \right|$.
$\Rightarrow {{A}_{1}}=\dfrac{1}{2}\left| \begin{matrix}
1 & 1 \\
9 & 1 \\
\end{matrix} \right|$.
$\Rightarrow {{A}_{1}}=\dfrac{1}{2}\left| \left( 1\times 1 \right)-\left( 1\times 9 \right) \right|$.
$\Rightarrow {{A}_{1}}=\dfrac{1}{2}\left| 1-9 \right|$.
$\Rightarrow {{A}_{1}}=\dfrac{1}{2}\left| -8 \right|$.
$\Rightarrow {{A}_{1}}=\dfrac{8}{2}$.
$\Rightarrow {{A}_{1}}=4$ units -(3).
From the figure, we can see that the line $x=c$ intersects the side AC and BC at points E and D respectively. Let us find these points.
Let us substitute $x=c$ in equation (1) to get the point E.
So, we have $y=\dfrac{c}{9}$. We get the point E as $\left( c,\dfrac{c}{9} \right)$ -(4).
Now, let us substitute $x=c$ in equation (2) to get the point D.
So, we have $y=1$. We get the point D as $\left( c,1 \right)$ -(5).
According to the problem, the line $x=c$ divides the area of the triangle into two equal halves. So, we get the area of the triangle ADE as $\dfrac{{{A}_{1}}}{2}=2$ units.
So, we have $2=\dfrac{1}{2}\left| \begin{matrix}
9-c & 1-1 \\
c-c & \dfrac{c}{9}-1 \\
\end{matrix} \right|$.
$\Rightarrow 4=\left| \begin{matrix}
9-c & 0 \\
0 & \dfrac{c-9}{9} \\
\end{matrix} \right|$.
$\Rightarrow 4=\left| \left( \left( 9-c \right)\times \left( \dfrac{c-9}{9} \right) \right)-\left( 0\times 0 \right) \right|$.
$\Rightarrow 4=\left| \dfrac{9c-81-{{c}^{2}}+9c}{9} \right|$.
$\Rightarrow 4=\left| \dfrac{+18c-81-{{c}^{2}}}{9} \right|$.
We know that ${{c}^{2}}>0$. So, we get $4=\dfrac{{{c}^{2}}-18c+81}{9}$.
$\Rightarrow 36={{\left( c-9 \right)}^{2}}$.
$\Rightarrow \left( c-9 \right)=\pm 6$.
$\Rightarrow \left( c-9 \right)=6$ or $\left( c-9 \right)=-6$.
$\Rightarrow c=15$ or $c=3$.
We neglect $c=15$ as all the x-coordinates of vertices of the triangle is less than or equal to 9.
So, we have found the value of c as 3.
∴ The value of c is 3.
Note: We can find the area of the triangle using different methods for example using heron’s formula. We can also find the area using the integration of the sides of the triangle between every two vertices. We should not consider the value of c as 15 as the line $x=15$ will not touch the sides of the triangle. We should not say the obtained answers without verifying it once with the given information of the problem. Similarly, we can also expect problems to find the angle between the sides and the line $x=c$.
We know that the equation of the line passing through two points $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ is $\left( y-{{y}_{1}} \right)=\dfrac{\left( {{y}_{2}}-{{y}_{1}} \right)}{\left( {{x}_{2}}-{{x}_{1}} \right)}\times \left( x-{{x}_{1}} \right)$.
We have the line segment AC passing through $A\left( 0,0 \right)$ and $C\left( 9,1 \right)$.
So, the equation of side AC is $\left( y-0 \right)=\dfrac{\left( 1-0 \right)}{\left( 9-0 \right)}\times \left( x-0 \right)$.
$\Rightarrow y=\dfrac{1}{9}\times x$.
$\Rightarrow y=\dfrac{x}{9}$ -(1).
We have the line segment BC passing through $B\left( 1,1 \right)$ and $C\left( 9,1 \right)$.
So, the equation of side BC is $\left( y-1 \right)=\dfrac{\left( 1-1 \right)}{\left( 9-1 \right)}\times \left( x-1 \right)$.
$\Rightarrow y-1=\dfrac{0}{8}\times \left( x-1 \right)$.
$\Rightarrow y=1$ -(2).
Now, let us find the area of the triangle ABC. Let us assume it as ${{A}_{1}}$.
We know that the area of the triangle with vertices $\left( {{x}_{1}},{{y}_{1}} \right)$, $\left( {{x}_{2}},{{y}_{2}} \right)$ and $\left( {{x}_{3}},{{y}_{3}} \right)$ is $\dfrac{1}{2}\left| \begin{matrix}
{{x}_{2}}-{{x}_{1}} & {{y}_{2}}-{{y}_{1}} \\
{{x}_{3}}-{{x}_{1}} & {{y}_{3}}-{{y}_{1}} \\
\end{matrix} \right|$.
$\Rightarrow {{A}_{1}}=\dfrac{1}{2}\left| \begin{matrix}
1-0 & 1-0 \\
9-0 & 1-0 \\
\end{matrix} \right|$.
$\Rightarrow {{A}_{1}}=\dfrac{1}{2}\left| \begin{matrix}
1 & 1 \\
9 & 1 \\
\end{matrix} \right|$.
$\Rightarrow {{A}_{1}}=\dfrac{1}{2}\left| \left( 1\times 1 \right)-\left( 1\times 9 \right) \right|$.
$\Rightarrow {{A}_{1}}=\dfrac{1}{2}\left| 1-9 \right|$.
$\Rightarrow {{A}_{1}}=\dfrac{1}{2}\left| -8 \right|$.
$\Rightarrow {{A}_{1}}=\dfrac{8}{2}$.
$\Rightarrow {{A}_{1}}=4$ units -(3).
From the figure, we can see that the line $x=c$ intersects the side AC and BC at points E and D respectively. Let us find these points.
Let us substitute $x=c$ in equation (1) to get the point E.
So, we have $y=\dfrac{c}{9}$. We get the point E as $\left( c,\dfrac{c}{9} \right)$ -(4).
Now, let us substitute $x=c$ in equation (2) to get the point D.
So, we have $y=1$. We get the point D as $\left( c,1 \right)$ -(5).
According to the problem, the line $x=c$ divides the area of the triangle into two equal halves. So, we get the area of the triangle ADE as $\dfrac{{{A}_{1}}}{2}=2$ units.
So, we have $2=\dfrac{1}{2}\left| \begin{matrix}
9-c & 1-1 \\
c-c & \dfrac{c}{9}-1 \\
\end{matrix} \right|$.
$\Rightarrow 4=\left| \begin{matrix}
9-c & 0 \\
0 & \dfrac{c-9}{9} \\
\end{matrix} \right|$.
$\Rightarrow 4=\left| \left( \left( 9-c \right)\times \left( \dfrac{c-9}{9} \right) \right)-\left( 0\times 0 \right) \right|$.
$\Rightarrow 4=\left| \dfrac{9c-81-{{c}^{2}}+9c}{9} \right|$.
$\Rightarrow 4=\left| \dfrac{+18c-81-{{c}^{2}}}{9} \right|$.
We know that ${{c}^{2}}>0$. So, we get $4=\dfrac{{{c}^{2}}-18c+81}{9}$.
$\Rightarrow 36={{\left( c-9 \right)}^{2}}$.
$\Rightarrow \left( c-9 \right)=\pm 6$.
$\Rightarrow \left( c-9 \right)=6$ or $\left( c-9 \right)=-6$.
$\Rightarrow c=15$ or $c=3$.
We neglect $c=15$ as all the x-coordinates of vertices of the triangle is less than or equal to 9.
So, we have found the value of c as 3.
∴ The value of c is 3.
Note: We can find the area of the triangle using different methods for example using heron’s formula. We can also find the area using the integration of the sides of the triangle between every two vertices. We should not consider the value of c as 15 as the line $x=15$ will not touch the sides of the triangle. We should not say the obtained answers without verifying it once with the given information of the problem. Similarly, we can also expect problems to find the angle between the sides and the line $x=c$.
Recently Updated Pages
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
What are the possible quantum number for the last outermost class 11 chemistry CBSE
Is C2 paramagnetic or diamagnetic class 11 chemistry CBSE
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Write an application to the principal requesting five class 10 english CBSE
Difference Between Plant Cell and Animal Cell
a Tabulate the differences in the characteristics of class 12 chemistry CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
Discuss what these phrases mean to you A a yellow wood class 9 english CBSE
List some examples of Rabi and Kharif crops class 8 biology CBSE