To solve the exercise we can first find the equation of a line perpendicular to the given lines.
Two lines are parallel if their lines are equal.
[tex]\begin{gathered} m_1=m_2 \\ \text{ Where }m_1\text{ is the slope of the first line and }m_2\text{ is the slope of the second line} \end{gathered}[/tex]And two lines are perpendicular if their respective slopes have the following relationship:
[tex]m_1=-\frac{1}{m_2}[/tex]Then, the slope of the line perpendicular to the given lines is:
[tex]\begin{gathered} m_1=-\frac{3}{2} \\ -\frac{3}{2}=-\frac{1}{m_2} \\ \text{ Multiply by -1 from both sides of the equation} \\ -1\cdot-\frac{3}{2}=-1\cdot-\frac{1}{m_2} \\ \frac{3}{2}=\frac{1}{m_2} \\ \text{ Apply the cross prduct} \\ 3\cdot m_2=1\cdot2 \\ 3m_2=2 \\ \text{ Divide by 3 from both sides of the equation} \\ \frac{3m_2}{3}=\frac{2}{3} \\ m_2=\frac{2}{3} \end{gathered}[/tex]Now, let us find a point through which the second line passes:
[tex]\begin{gathered} x=0 \\ y=-\frac{3}{2}x-1\Rightarrow\text{ Second line} \\ \text{ Replace the value of x} \\ y=-\frac{3}{2}(0)-1 \\ y=0-1 \\ y=-1 \\ \text{ Then, this line passes through the point (0,-1)} \end{gathered}[/tex]Now, using the point-slope formula we can find the equation of the line perpendicular to the given parallel lines:
[tex]\begin{gathered} $y-y_1=m(x-x_1)$\Rightarrow\text{ Point-slope formula} \\ \text{ Where m is the slope of the line and }(x_1,y_1)\text{ is a point through the line passes} \end{gathered}[/tex][tex]\begin{gathered} m=\frac{2}{3} \\ (x_1,y_1)=(0,1) \\ y-y_1=m(x-x_1) \\ y-(-1)=\frac{2}{3}(x-0) \\ y-(-1)=\frac{2}{3}(x) \\ y+1=\frac{2}{3}x \\ \text{ Subtract -1 from both sides of the equation} \\ y+1-1=\frac{2}{3}x-1 \\ y+1-1=\frac{2}{3}x-1 \\ $\boldsymbol{y=\frac{2}{3}x-1}$ \end{gathered}[/tex]Now, we can find the point where the first line and the line that is perpendicular to the two given lines meet:
[tex]\begin{gathered} y=-\frac{3}{2}x+4\Rightarrow\text{ First line } \\ y=\frac{2}{3}x-1\Rightarrow\text{ Perpendicular line} \\ \text{ We equalize and solve for x} \\ -\frac{3}{2}x+4=\frac{2}{3}x-1 \\ \text{ Subtract 4 from both sides of the equation} \\ -\frac{3}{2}x+4-4=\frac{2}{3}x-1-4 \\ -\frac{3}{2}x=\frac{2}{3}x-5 \\ \text{ Subtract }\frac{2}{3}x\text{ from both sides of the equation} \\ -\frac{3}{2}x-\frac{2}{3}x=\frac{2}{3}x-5-\frac{2}{3}x \\ (-\frac{3}{2}-\frac{2}{3})x=-5 \\ (\frac{-3\cdot3-2\cdot2}{2\cdot3})x=-5 \\ (\frac{-9-4}{6})x=-5 \\ \frac{-13}{6}x=-5 \\ \text{Multiply by 6 from both sides of the equation} \\ 6\cdot\frac{-13}{6}x=-5\cdot6 \\ -13x=-30 \\ \text{ Divide by -13 from both sides of the equation} \\ \frac{-13x}{-13}=\frac{-30}{-13} \\ x=\frac{30}{13} \\ x\approx2.3\Rightarrow\approx\text{ it reads "approximately"} \end{gathered}[/tex]Now, we can plug the value of x into the equation of the first line to find its respective y-coordinate:
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