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**Hint:**Firstly, we have to draw a figure with the given data. Then, we have to find the equation of the line AB, BC and AC using the formula for a line joining two points $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ which is given by $y-{{y}_{1}}=\left( \dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \right)\left( x-{{x}_{1}} \right)$ . Then, we have to check whether the given point point is placed above or below AB, BC and AC since the point is present inside the triangle ABC. Accordingly, we will create a linear inequality using the equations of lines AB, BC and AC. If the point is below a line gegment, then the equation of the corresponding line segment will be less than 0, at the same time, we have to substitute the coordinates of the point in the equation. If the point is above the line, then the equation of the corresponding line will be greater than 0. Here also, we have to substitute the coordinates of the point in the equation. Then, we have to find the value of t and reach a conclusion on its value by looking into the three values of t obtained from the above method.

**Complete step-by-step solution:**

We have to find the values of t such that the point $\left( t,t+1 \right)$ lies inside the triangle ABC where \[A=\left( 0,3 \right),B=\left( -2,0 \right)\text{ and }C=\left( 6,1 \right)\] . Let us draw a triangle with the given values.

We have to find the equation of the line AB. We know that equation of a line with points $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ is given by

$y-{{y}_{1}}=\left( \dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \right)\left( x-{{x}_{1}} \right)...\left( i \right)$

We are given with the points of line segment AB as \[A=\left( 0,3 \right)\text{ and }B=\left( -2,0 \right)\] . Therefore, equation of the line AB can be found by substituting \[{{x}_{1}}=0,{{x}_{2}}=-2,{{y}_{1}}=3\text{ and }{{y}_{2}}=0\] in the equation (i).

$\begin{align}

& \Rightarrow y-3=\dfrac{0-3}{-2-0}\left( x-0 \right) \\

& \Rightarrow y-3=\dfrac{-3}{-2}x \\

& \Rightarrow y-3=\dfrac{3}{2}x \\

\end{align}$

Let us take the 2 in the denominator of the RHS to the LHS.

$\Rightarrow 2\left( y-3 \right)=3x$

We have to apply the distributive law in the LHS.

$\Rightarrow 2y-6=3x$

Let us take the terms in the LHS to the RHS.

$\Rightarrow 3x-2y+6=0...\left( ii \right)$

Similarly, we have to find the equation of the line AC. We are given the points \[A=\left( 0,3 \right)\] and \[C=\left( 6,1 \right)\] .

$\begin{align}

& \Rightarrow y-3=\dfrac{1-3}{6-0}\left( x-0 \right) \\

& \Rightarrow y-3=\dfrac{-2}{6}x \\

& \Rightarrow y-3=\dfrac{-1}{3}x \\

\end{align}$

Let us take the coefficient of x to the RHS.

$\Rightarrow -3\left( y-3 \right)=x$

We have to apply distributive property on the LHS.

$\begin{align}

& \Rightarrow -3\times y-\left( -3 \right)\times 3=x \\

& \Rightarrow -3y+9=x \\

\end{align}$

Let us take the terms in the LHS to the RHS.

$\Rightarrow x+3y-9=0...\left( iii \right)$

Now, let us find the equation of the line BC. We are given the points \[B=\left( -2,0 \right)\text{ and }C=\left( 6,1 \right)\] .

$\begin{align}

& \Rightarrow y-0=\dfrac{1-0}{6-\left( -2 \right)}\left( x-\left( -2 \right) \right) \\

& \Rightarrow y=\dfrac{1}{6+2}\left( x+2 \right) \\

& \Rightarrow y=\dfrac{1}{8}\left( x+2 \right) \\

\end{align}$

Let us take $\dfrac{1}{8}$ to the LHS.

$\Rightarrow 8y=x+2$

We have to take 8y to the RHS.

$\Rightarrow x-8y+2=0...\left( iv \right)$

Let us consider an equation of a line $ax+by+c=0$ and a point $\left( {{x}_{1}},{{y}_{1}} \right)$ , then $a{{x}_{1}}+b{{y}_{1}}+c<0$ if the point $\left( {{x}_{1}},{{y}_{1}} \right)$ is below the line and $a{{x}_{1}}+b{{y}_{1}}+c>0$ if the point $\left( {{x}_{1}},{{y}_{1}} \right)$ is above the line.

So let us consider the point $\left( t,t+1 \right)$ . We can see that this point is below the line AB Therefore, we will consider the inequality $a{{x}_{1}}+b{{y}_{1}}+c<0$ . Let us substitute $x=t$ and $y=t+1$ in the equation (ii).

$\begin{align}

& \Rightarrow 3t-2\left( t+1 \right)+6<0 \\

& \Rightarrow 3t-2t-2+6<0 \\

& \Rightarrow t+4<0 \\

\end{align}$

Let us take 4 to the RHS.

$\Rightarrow t<-4...\left( v \right)$

Now, we have to consider the point $\left( t,t+1 \right)$ which is below the line AC. Let us substitute $x=t$ and $y=t+1$ in the equation (iii).

$\begin{align}

& \Rightarrow t+3\left( t+1 \right)-9<0 \\

& \Rightarrow t+3t+3-9<0 \\

& \Rightarrow 4t-6<0 \\

\end{align}$

Let us take -6 to the RHS.

$\Rightarrow 4t<6$

We have to take 4 to the RHS.

$\begin{align}

& \Rightarrow t<\dfrac{6}{4} \\

& \Rightarrow t<\dfrac{3}{2}...\left( vi \right) \\

\end{align}$

Now, let us consider the point $\left( t,t+1 \right)$ which is above the line BC. Therefore, we will consider the inequality $a{{x}_{1}}+b{{y}_{1}}+c>0$ . Let us substitute $x=t$ and $y=t+1$ in the equation (iv).

$\begin{align}

& \Rightarrow t-8\left( t+1 \right)+2>0 \\

& \Rightarrow t-8t-8+2>0 \\

& \Rightarrow -7t-6>0 \\

\end{align}$

Let us take -6 to the RHS.

$\begin{align}

& \Rightarrow -7t>0+6 \\

& \Rightarrow -7t>6 \\

\end{align}$

We have to take the coefficient of t to the RHS.

$\Rightarrow t<-\dfrac{6}{7}..\left( vii \right)$

From (v), (vi) and (vii), we can write $t<-4$ or $t\in \left( -\infty , -4 \right)$ .

**Hence, the value of t must be such that $t\in \left( -\infty , -4 \right)$ .**

**Note:**Students must be through with the general formula of equation of a line joining two points. They must carefully put the appropriate inequality to find the value of t. They must note that we used open brackets in the final value of t because the inequality sign is not a ‘greater than or equal to’ sign or ‘less than or equal to’ sign.

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