What is the equation of a line that passes through the point (8, 1) and is perpendicular to the line whose equation is y = −2/3x + 5?
+9466 The slope of the line y = -\(\frac23\)x + 5 is -\(\frac23\)
And the slope of a perpendicular line is the negative reciprocal of - \(\frac23\) , which is \(\frac32\) .
Using the point (8, 1) and the slope \(\frac32\) , the equation of our line in point-slope form is...
y - 1 = \(\frac32\)(x - 8) This equation is correct as is, but if you need it in slope-intercept form,
distribute the \(\frac32\) to both terms in parenthesees.
y - 1 = \(\frac32\)x - 12
Add 1 to both sides of the equation.
y = \(\frac32\)x - 11
+9466 The slope of the line y = -\(\frac23\)x + 5 is -\(\frac23\)
And the slope of a perpendicular line is the negative reciprocal of - \(\frac23\) , which is \(\frac32\) .
Using the point (8, 1) and the slope \(\frac32\) , the equation of our line in point-slope form is...
y - 1 = \(\frac32\)(x - 8) This equation is correct as is, but if you need it in slope-intercept form,
distribute the \(\frac32\) to both terms in parenthesees.
y - 1 = \(\frac32\)x - 12
Add 1 to both sides of the equation.
y = \(\frac32\)x - 11
