SOLUTION:
Step 1:
In this question, we are given the following:
Which equation represents the line passing through the point (-5, 12) and perpendicular to the line y=2/3x +14 ?
Step 2:
The details of the solution are as follows:
[tex]\begin{gathered} Given\text{ that the equation is given as:} \\ \text{y = }\frac{2x}{3}\text{ + 14} \\ compari\text{ng this with:} \\ \text{y = mx + c , we have that:} \\ \text{m}_1\text{ =}\frac{2}{3} \end{gathered}[/tex][tex]\begin{gathered} Now,\text{ for perpendicular lines, we have that:} \\ m_1m_2\text{ = -1} \\ Then,\text{ we have that:} \\ m_2\text{ = }\frac{-1}{m_1}\text{ , where m}_1\text{ = }\frac{2}{3} \\ Then, \\ m_2\text{ = -1 }\div\text{ }\frac{2}{3} \\ m_2=\text{ -1 x }\frac{3}{2} \\ m_2\text{ = }\frac{-3}{2} \end{gathered}[/tex]Step 3:
[tex]\begin{gathered} Given\text{ that:} \\ m_2=\frac{-3}{2} \\ (\text{ x}_{1,}y_1)\text{ = \lparen -5, 12 \rparen} \end{gathered}[/tex][tex]\begin{gathered} Using\text{ the new equation:} \\ y\text{ -y}_1\text{ = m \lparen x - x}_1) \\ y\text{ - \lparen12\rparen = }\frac{-3}{2}\text{ \lparen x --5\rparen} \\ \text{y -12 = }\frac{-3}{2}(\text{ x+ 5 \rparen} \\ Multiply\text{ through by 2, we have that:} \\ 2y\text{ - 24 = - 3 \lparen x + 5\rparen} \\ 2y\text{ - 24 = -3x - 15} \\ Re-arranging,\text{ we have that:} \\ 3x+\text{ 2y - 24 + 15 = 0} \\ 3x+\text{ 2y -9 = 0} \end{gathered}[/tex]CONCLUSION:
The final answer is:
[tex]3x\text{ + 2y -9 = 0}[/tex]
