Respuesta :
Answer:
The point-slope form of the line that passes through (6,1) and is parallel to a line with a slope of -3 is 3x + y – 19 = 0
Solution:
The point slope form of the line that passes through the points [tex]\left(x_{1} y_{1}\right)[/tex] and parallel to the line with slope “m” is given as
[tex]y - y_{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” = -3
Given that the line passes through the points (6,1).Hence we get [tex]x_{1} = 6 ; y_{1} = 1[/tex]
By substituting the values in eqn 1, we get the point slope form of the line which is parallel to the line having slope -3 can be found out.
y – 1 = -3(x – 6)
y – 1 = -3x +18
On rearranging the terms, we get
3x + y -1 – 18 = 0
3x + y – 19 = 0
Hence the point slope form of given line is 3x + y – 19 = 0
