Respuesta :
Since the line we want if perpendicular to 5x-4y=10, we first want to know the slope of this line. To do that, let's rewrite it in the slope-intercept form:
[tex]\begin{gathered} 5x-4y=10 \\ 4y=5x-10 \\ y=\frac{5x-10}{4}=\frac{5}{4}x-\frac{10}{4} \\ y=\frac{5}{4}x-\frac{5}{2} \end{gathered}[/tex]The slope of this line is the linear factor, the one multiplying x: 5/4.
To get the slope of a perpendicular line, we invert it and chenge its sign:
[tex]s_{\text{perpendicular}}=-\frac{1}{s}=-\frac{4}{5}[/tex]So, we have, so far:
[tex]y=-\frac{4}{5}x+b[/tex]We need to find b, and for this we use the point (10,6):
[tex]\begin{gathered} 6=-\frac{4}{5}\cdot10+b \\ 6=-4\cdot2+b \\ 6=-8+b \\ b=6+8 \\ b=14 \end{gathered}[/tex]So, the euqation of perpedicular line that passes through (10, 6) is:
[tex]y=-\frac{4}{5}x+14[/tex]