Solution:
To find the equation of a line that passes through the point (2,3) and perpendicular to 2x + 3y =4, in slope-intercept form.
Concept:
Two lines are perpendicular if their slopes are negative reciprocals of each other.
[tex]\begin{gathered} m_1m_2=-1 \\ \text{where;} \\ m_1\text{ is the slope of line 1} \\ m_2\text{ is the slope of line }2 \end{gathered}[/tex]Given:
Line 1
[tex]\begin{gathered} 2x+3y=4 \\ 3y=-2x+4 \\ \text{Dividing through by 3 to make the equation in slope-intercept form;} \\ y=-\frac{2}{3}x+\frac{4}{3} \\ \\ \text{Comparing to the equation of a line in slope-intercept form,} \\ y=mx+C \\ \text{where m is the slope} \\ C\text{ is the y-intercept} \\ y=mx+C \\ y=-\frac{2}{3}x+\frac{4}{3} \\ \text{Then the slope of line 1 is;} \\ m_1=-\frac{2}{3} \end{gathered}[/tex]Line 2
Since line 2 is perpendicular to line 1, then;
[tex]m_1m_2=-1[/tex]The slope of line 2 is calculated below;
[tex]\begin{gathered} m_1m_2=-1 \\ -\frac{2}{3}\times m_2=-1 \\ m_2=-1\times\frac{-3}{2} \\ m_2=\frac{3}{2} \end{gathered}[/tex]The equation of line 2 in slope-intercept form passing through the point (2,3) is;
[tex]\begin{gathered} y-y_1=m(x-x_1) \\ \text{where;} \\ m=\frac{3}{2} \\ x_1=2 \\ y_1=3 \\ \\ \text{Substituting these into the equation,} \\ y-3=\frac{3}{2}(x-2) \\ \text{Cross multiplying the equation,} \\ 2(y-3)=3(x-2) \\ \text{Expanding the brackets,} \\ 2y-6=3x-6 \\ \text{Collecting the like terms,} \\ 2y=3x-6+6 \\ 2y=3x+0 \\ Divid\text{ ing both sides by 2 to leave the equation in slope-intercept form,} \\ y=\frac{3}{2}x+0c \\ y=\frac{3}{2}x \end{gathered}[/tex]Therefore, the line that passes through the point (2,3) and is perpendicular to the line 2x + 3y = 4 in slope-intercept form is;
[tex]\begin{gathered} y=\frac{3}{2}x+0c \\ y=\frac{3}{2}x \end{gathered}[/tex]
