Respuesta :
Answer:
The point-slope form of the line that passes through (5,5) and is perpendicular to a line with a slope of [tex]\frac{1}{4}[/tex] is 4x + y -25 = 0
Solution:
The point slope form of the line that passes through the points [tex]\left(x_{1} y_{1}\right)[/tex] and perpendicular to the line with a slope of “m” is given as
[tex]\bold{y-y_{1}=-\frac{1}{m}\left(x-x_{1}\right)}[/tex] ---- eqn 1
Where “m” is the slope of the line. [tex]x_{1} \text { and } y_{1}[/tex] are the points that passes through the line.
From question, given that slope “m” = [tex]\frac{1}{4}[/tex]
Given that the line passes through the points (5,5).Hence we get
[tex]x_{1}=5 ; y_{1}=5[/tex]
By substituting the values in eqn 1 , we get the point slope form of the line which is perpendicular to the line having slope [tex]\frac{1}{4}[/tex]can be found out.
y - 5 = -4(x - 5)
y - 5 = -4x + 20
on simplifying the above equation, we get
y - 5 + 4x -20 = 0
4x + y - 25 = 0
hence the point slope form of given line is 4x + y - 25 = 0
