Question

Could you please help me with #9 I don't understand how to do it

Answer 1

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.

`\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}`

Given:

Line 1

`\begin{gathered} 2x+3y=4 \\ 3y=-2x+4 \\ \text{Dividing through by 3 to make the equation in slope-intercept form;} \\ y=-(2)/(3)x+(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=-(2)/(3)x+(4)/(3) \\ \text{Then the slope of line 1 is;} \\ m_1=-(2)/(3) \end{gathered}`

Line 2

Since line 2 is perpendicular to line 1, then;

`m_1m_2=-1`

The slope of line 2 is calculated below;

`\begin{gathered} m_1m_2=-1 \\ -(2)/(3)* m_2=-1 \\ m_2=-1*(-3)/(2) \\ m_2=(3)/(2) \end{gathered}`

The equation of line 2 in slope-intercept form passing through the point (2,3) is;

`\begin{gathered} y-y_1=m(x-x_1) \\ \text{where;} \\ m=(3)/(2) \\ x_1=2 \\ y_1=3 \\ \\ \text{Substituting these into the equation,} \\ y-3=(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=(3)/(2)x+0c \\ y=(3)/(2)x \end{gathered}`

Therefore, the line that passes through the point (2,3) and is perpendicular to the line 2x + 3y = 4 in slope-intercept form is;

`\begin{gathered} y=(3)/(2)x+0c \\ y=(3)/(2)x \end{gathered}`