Answer:
Equation of a line through (15, 8) that is perpendicular to the line y =57 x −1 is [tex]\mathbf{y=-\frac{1}{57}x+\frac{157}{19}}[/tex]
Step-by-step explanation:
We need to write an equation of a line through (15, 8) that is perpendicular to the line y =57 x −1.
The equation will be in slope-intercept form: [tex]\mathbf{y=mx+b}[/tex]
Where m is slope and b is y-intercept.
Finding Slope of new line
We are given equation [tex]y =57x-1[/tex] of line, which is perpendicular to the line whose equation we need to find.
If the lines are perpendicular there slopes are inverse of each other i.e [tex]m_1=-\frac{1}{m_2}[/tex]
Slope of given line is m= 57 ( compare the given equation with standard slope-intercept form the value of m is 57
Slope of new line is m= [tex]-\frac{1}{57}[/tex]
Finding y-intercept of new line
Now finding y-intercept (b) for new line. It can be found using point (15,8) and slope m = [tex]-\frac{1}{57}[/tex]
[tex]y=mx+b\\8=-\frac{1}{57}(15)+b\\8=-\frac{15}{57}+b\\b=8+\frac{15}{57}\\b=\frac{8*57+15}{57}\\b=\frac{471}{57}\\b=\frac{157}{19}[/tex]
So, y-intercept b of new line is: [tex]b=\frac{157}{19}[/tex]
Equation of required line
Equation of line having slope m= [tex]-\frac{1}{57}[/tex] and y-intercept [tex]b=\frac{157}{19}[/tex] is:
[tex]y=mx+b\\\mathbf{y=-\frac{1}{57}x+\frac{157}{19}}[/tex]
So, equation of a line through (15, 8) that is perpendicular to the line y =57 x −1 is [tex]\mathbf{y=-\frac{1}{57}x+\frac{157}{19}}[/tex]
