Question

The
area bounded
bounded by the
lines |2x-3| +|3y+4|=5 is​

Answer 1

Answer:

This problem is not difficult, but tedious. Here is what needs to be done.

Consider four cases

2x-3>0, 3y-2>0 then (2x-3)+(3y-2) =6 or y=-(2/3)x+11/3

2x-3>0, 3y-2<0 then (2x-3)-(3y-2) =6 or y= (2/3)x +7/3

2x-3<0, 3y-2>0 then -(2x-3)+(3y-2) =6 or y =(2/3)x+11/3

2x-3<0, 3y-2<0 then -(2x-3)-(3y-2) =6 or y= -(2/3)x + 1/3.

Hope it helps u dear

Answer 2

The lines are given by the equation,

$\longrightarrow|2x-3|+|3y+4|=5$

or,

$\longrightarrow|3y+4|=5-|2x-3|$

Case 1:-

Let $x\geq\dfrac{3}{2}\ \implies\ |2x-3|=2x-3.$

Case 1.1:-

Let $y\geq-\dfrac{4}{3}\ \implies\ |3y+4|=3y+4.$

Then the equation becomes,

$\longrightarrow3y+4=5-2x+3$

$\longrightarrow3y+4=8-2x$

$\longrightarrow\underline{y=\dfrac{4-2x}{3}}$

Case 1.2:-

Let $y\leq-\dfrac{4}{3}\ \implies\ |3y+4|=-3y-4.$

Then the equation becomes,

$\longrightarrow-3y-4=5-2x+3$

$\longrightarrow3y+4=2x-8$

$\longrightarrow\underline{y=\dfrac{2x-12}{3}}$

We need to find the point of intersection of these two lines.

Equating them,

$\longrightarrow \dfrac{4-2x}{3}=\dfrac{2x-12}{3}$

$\longrightarrow x=4$

So the area bounded by these two lines, from $x=\dfrac{3}{2}$ to $x=4$ is,

$\displaystyle\longrightarrow A_1=\int\limits_{\frac{3}{2}}^4\left(\dfrac{4-2x}{3}-\dfrac{2x-12}{3}\right)\ dx$

$\displaystyle\longrightarrow A_1=\dfrac{1}{3}\int\limits_{\frac{3}{2}}^4\left(16-4x\right)\ dx$

$\displaystyle\longrightarrow A_1=\dfrac{1}{3}\left[16\big[x\big]_{\frac{3}{2}}^4-2\left[x^2\right]_{\frac{3}{2}}^4\right]$

$\displaystyle\longrightarrow A_1=\dfrac{1}{3}\left[16\cdot\dfrac{5}{2}-2\cdot\dfrac{55}{4}\right]$

$\displaystyle\longrightarrow A_1=\dfrac{25}{6}$

Case 2:-

Let $x\leq\dfrac{3}{2}\ \implies\ |2x-3|=3-2x.$

Case 2.1:-

Let $y\geq-\dfrac{4}{3}\ \implies\ |3y+4|=3y+4.$

Then the equation becomes,

$\longrightarrow3y+4=5+2x-3$

$\longrightarrow3y+4=2x+2$

$\longrightarrow\underline{y=\dfrac{2x-2}{3}}$

Case 2.2:-

Let $y\leq-\dfrac{4}{3}\ \implies\ |3y+4|=-3y-4.$

Then the equation becomes,

$\longrightarrow-3y-4=5+2x-3$

$\longrightarrow3y+4=-2x-2$

$\longrightarrow\underline{y=\dfrac{-2x-6}{3}}$

We need to find the point of intersection of these two lines.

Equating them,

$\longrightarrow\dfrac{2x-2}{3}=\dfrac{-2x-6}{3}$

$\longrightarrow x=-1$

So the area bounded by these two lines, from $x=-1$ to $x=\dfrac{3}{2}$ is,

$\displaystyle\longrightarrow A_2=\int\limits_{-1}^{\frac{3}{2}}\left(\dfrac{2x-2}{3}-\dfrac{-2x-6}{3}\right)\ dx$

$\displaystyle\longrightarrow A_2=\dfrac{1}{3}\int\limits_{-1}^{\frac{3}{2}}\left(4x+4\right)\ dx$

$\displaystyle\longrightarrow A_2=\dfrac{1}{3}\left[2\left[x^2\right]_{-1}^{\frac{3}{2}}+4\big[x\big]_{-1}^{\frac{3}{2}}\right]$

$\displaystyle\longrightarrow A_2=\dfrac{1}{3}\left[2\cdot\dfrac{5}{4}+4\cdot\dfrac{5}{2}\right]$

$\displaystyle\longrightarrow A_2=\dfrac{25}{6}$

Hence the total area is,

$\longrightarrow A=A_1+A_2$

$\displaystyle\longrightarrow A=\dfrac{25}{6}+\dfrac{25}{6}$

$\displaystyle\longrightarrow\underline{\underline{A=\dfrac{25}{3}}}$