Respuesta :
The equation of the line that is perpendicular to 4x - 3y = 10 through the point (-2,4) is [tex]y-4=\frac{-3}{4}(x+2)[/tex]
Solution:
Given, line equation is 4x – 3y = 10
We have to find a line that is perpendicular to 4x – 3y = 10 and passing through (-2, 4)
Now, let us find the slope of the given line,
[tex]\text { Slope of a line }=\frac{-\mathrm{x} \text { coefficient }}{\mathrm{y} \text { coefficient }}=\frac{-4}{-3}=\frac{4}{3}[/tex]
We know that, slope of a line [tex]\times[/tex] slope of perpendicular line = -1
[tex]\begin{array}{l}{\text { Then, } \frac{4}{3} \times \text { slope of perpendicular line }=-1} \\\\ {\rightarrow \text { slope of perpendicular line }=-1 \times \frac{3}{4}=-\frac{3}{4}}\end{array}[/tex]
Now, slope of our required line = [tex]\frac{-3}{4}[/tex] and it passes through (-2, 4)
The point slope form is given as:
[tex]\begin{array}{l}{y-y_{1}=m\left(x-x_{1}\right) \text { where } m \text { is slope and }\left(x_{1}, y_{1}\right) \text { is point on the line. }} \\\\ {\text { Here in our problem, } m=-\frac{3}{4}, \text { and }\left(x_{1}, y_{1}\right)=(-2,4)} \\\\ {\text { Then, line equation } \rightarrow y-4=-\frac{3}{4}(x-(-2))}\end{array}[/tex]
[tex]y-4=\frac{-3}{4}(x+2)[/tex]
Hence the equation of line is found out
