Answer:
[tex]2x-5y+7=0[/tex]
Step-by-step explanation:
The altitude of a triangle is a perpendicular line segment drawn from a vertex to its opposite side.
To find the equation of the altitude of triangle MNP drawn from point N, we first need to determine the slope of the line opposite vertex N, which is the slope of side MP.
[tex]\boxed{\begin{array}{l}\underline{\textsf{Slope formula}}\\\\m=\dfrac{y_2-y_1}{x_2-x_1}\\\\\textsf{where:}\\\phantom{w}\bullet\;\;m\; \textsf{is the slope.}\\\phantom{w}\bullet\;\;(x_1,y_1)\;\textsf{and}\;(x_2,y_2)\;\textsf{are two points on the line.}\end{array}}[/tex]
To find the slope of side MP, substitute points M(3, 4) and P(5, -1) into the slope formula:
[tex]m_{MP}=\dfrac{y_P-y_M}{x_P-x_M}=\dfrac{-1-4}{5-3}=-\dfrac{5}{2}[/tex]
Since the altitude is perpendicular to side MP, the negative reciprocal of the slope of side MP gives the slope of the altitude. So, the slope of the altitude is:
[tex]m=\dfrac{-1}{-\frac{5}{2}}=\dfrac{2}{5}[/tex]
[tex]\boxed{\begin{array}{l}\underline{\textsf{Point-slope form of a linear equation}}\\\\y-y_1=m(x-x_1)\\\\\textsf{where:}\\ \phantom{ww}\bullet\;\textsf{$m$ is the slope.}\\\phantom{ww}\bullet\;\textsf{ $(x_1,y_1)$ is a point on the line.}\end{array}}[/tex]
To write the equation of the altitude, substitute the slope (m = 2/5) and point N(-1, 1) into the point-slope form of a linear equation:
[tex]\begin{aligned}y-y_N&=m(x-x_N)\\\\y-1&=\dfrac{2}{5}(x-(-1))\\\\5y-5&=2(x+1)\\\\5y-5&=2x+2\\\\2x-5y+2+5&=0\\\\2x-5y+7&=0\end{aligned}[/tex]
Therefore, the equation of the altitude of triangle MNP drawn from point N is:
[tex]\Large\boxed{\boxed{2x-5y+7=0}}[/tex]
