# The line $x=c$ cuts the triangle with corners $\left( 0,0 \right)$, $\left( 1,1 \right)$ and $\left( 9,1 \right)$ into two regions. If the area of the two regions is the same, then find the value of c?

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**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.

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.

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

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