Respuesta :
a) y = 3/5x + 11/5
b) y = -5/3x + 9
Explanation:[tex]\begin{gathered} a)\text{ }y\text{ = }\frac{3}{5}x\text{ - 3} \\ \text{compare with equation of line:} \\ y\text{ = mx + b} \\ m\text{ =slope, b = y-intercept} \\ m\text{ =slope = 3/5} \\ b\text{ = -3} \end{gathered}[/tex]For a line to be parallel to another line. the slope of the 1st line will be equalt to the slope of the 2nd line:
slope of 1st line = 3/5
So, the slope of the 2nd line = 3/5
Given point: (3, 4) = (x, y)
To get the y-intercept of the second line, we would insert the slope and the point into the equation of line
[tex]\begin{gathered} y\text{ = mx + b} \\ 4\text{ = }\frac{3}{5}(3)\text{ + b} \\ 4\text{ = 9/5 + b} \\ 4\text{ - }\frac{\text{9}}{5}\text{ = b} \\ \frac{20-9}{5}\text{ = b} \\ b\text{ = 11/5} \end{gathered}[/tex]The equation of line parallel to y = 3/5x - 3:
[tex]\begin{gathered} y\text{ = mx + b} \\ y\text{ = }\frac{3}{5}x\text{ + }\frac{11}{5} \end{gathered}[/tex][tex]b)\text{ line perpendicular to y = 3/5x - 3}[/tex]For a line to be perpendicular to another line, the slope of one will be the negative reciprocal of the second line
Slope of the 1st line = 3/5
reciprocal of 3/5 = 5/3
negative reciprocal = -5/3
slope of the 2nd line (perpendicular) = -5/3
We need to get the y-intercept of the perpendicular line:
[tex]\begin{gathered} \text{given point: (3,4) = (x, y)} \\ y\text{ = mx + b} \\ m\text{ of the perpendicular = -5/3} \\ 4\text{ = }\frac{-5}{3}(3)\text{ + b} \\ 4\text{ = -5 + b} \\ 4\text{ + 5 = b} \\ b\text{ = 9} \end{gathered}[/tex]The equation of line perpendicular to y = 3/5x - 3:
[tex]\begin{gathered} y\text{ = mx + b} \\ y\text{ = }\frac{-5}{3}x\text{ + 9} \end{gathered}[/tex]