Respuesta :
Answer:
1) The equation of the line parallel to x-5y=6 and through (4,-2) is 5y = x -14
2) The equation of the line perpendicular to y= -2/5x + 3 and through (2,-1) is 2y = 5x -12
Solution:
1) find the equation of the line parallel to x-5y=6 and through (4,-2).
Given, line equation is x – 5y = 6
We have to find the line equation that is parallel to given line and passing through the point (4, -2)
Now, let us find slope of the given line.
[tex]\text { Slope }=\frac{-x \text { coefficient }}{y \text { coefficient }}=\frac{-1}{-5}=\frac{1}{5}[/tex]
Now, we know that, slope of parallel lines are equal.
So, slope of required line is 1/5 and it passes through (4, -2)
Now, using point slope form
[tex]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 line. }[/tex]
[tex]y-(-2)=\frac{1}{5}(x-4) \rightarrow 5(y+2)=x-4 \rightarrow 5 y+10=x-4 \rightarrow x-5 y=14[/tex]
Hence, the line equation is 5y = x -14
2) find the equation of the line perpendicular to y= -2/5x + 3 and through (2,-1)
[tex]\text { Given, line equation is } y=-\frac{2}{5} x+3 \rightarrow 5 y=-2 x+15 \rightarrow 2 x+5 y=15[/tex]
We have to find the line equation that is perpendicular to given line and passing through the point (2, -1)
Now, let us find slope of the given line.
[tex]\text { Slope }=\frac{-x \text { coefficient }}{y \text { coefficient }}=\frac{-2}{5}=-\frac{2}{5}[/tex]
Now, we know that, product of slopes of perpendicular lines equals to -1.
So, slope of required line [tex]\times[/tex] slope of given line = -1
slope of required line = [tex]-1 \times \frac{5}{-2}=\frac{5}{2}[/tex]
And it passes through (2, -1)
Now, using point slope form
[tex]\begin{array}{l}{\text { Line equation is } y-(-1)=\frac{5}{2}(x-2) \rightarrow 2(y+1)=5(x-2)} \\\\ {\rightarrow 2 y+2=5 x-10 \rightarrow 5 x-2 y=12}\end{array}[/tex]
Hence, the line equation is 2y = 5x -12
