Resourcesdown chevron
Testimonials
Plans
Sign in
Sign up
Resourcesright arrow
Testimonials
Plans
Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

The equation of line qq is y=16x−5y = \frac{1}{6}x - 5. Line rr, which is parallel to line qq, includes the point (6,2)(6,2). What is the equation of line rr?\newlineWrite the equation in slope-intercept form. Write the numbers in the equation as simplified proper fractions, improper fractions, or integers.

View step-by-step helpView step-by-step help
Homeright chevron icon
Math Problemsright chevron icon
Algebra 1right chevron icon
Write an equation for a parallel or perpendicular lineright chevron icon

Full solution

Q. The equation of line qq is y=16x−5y = \frac{1}{6}x - 5. Line rr, which is parallel to line qq, includes the point (6,2)(6,2). What is the equation of line rr?\newlineWrite the equation in slope-intercept form. Write the numbers in the equation as simplified proper fractions, improper fractions, or integers.
  1. Find slope of line q: Determine the slope of line q. Line q has the equation y=16x−5y = \frac{1}{6}x - 5. The slope of line q is the coefficient of xx, which is 16\frac{1}{6}.
  2. Determine slope of line rr: Since line rr is parallel to line qq, it must have the same slope. The slope of line rr is therefore also 16\frac{1}{6}.
  3. Use point-slope form: Use the point-slope form to find the equation of line rr. The point-slope form of a line is y−y1=m(x−x1)y - y_1 = m(x - x_1), where mm is the slope and (x1,y1)(x_1, y_1) is a point on the line. We have the point (6,2)(6,2) and the slope 16\frac{1}{6}.
  4. Substitute slope and point: Substitute the slope and the point into the point-slope form.\newlineUsing the point (6,2)(6,2) and the slope 16\frac{1}{6}, the equation becomes y−2=16(x−6)y - 2 = \frac{1}{6}(x - 6).
  5. Simplify equation: Simplify the equation to get it into slope-intercept form, y=mx+by = mx + b.\newliney−2=16x−16(6)y - 2 = \frac{1}{6}x - \frac{1}{6}(6)\newliney−2=16x−1y - 2 = \frac{1}{6}x - 1\newliney=16x−1+2y = \frac{1}{6}x - 1 + 2\newliney=16x+1y = \frac{1}{6}x + 1

More problems from Write an equation for a parallel or perpendicular line

Question
Does the point (0,4)(0, 4) satisfy the equation y=xy = x??\newlineChoices: [[yes]][[no]]
Get tutor helpright-arrow

Posted 2 months ago

Question
Find the slope of the line y=−4x+25 y = -4x + \frac{2}{5} .\newlineWrite your answer as an integer or as a simplified proper or improper fraction.
Get tutor helpright-arrow

Posted 1 month ago

Question
Rewrite the following equation in slope-intercept form.\newliney+8=−8(x−10)y + 8 = -8(x - 10)\newlineWrite your answer using integers, proper fractions, and improper fractions in simplest form.\newline_____
Get tutor helpright-arrow

Posted 1 month ago

Question
Find the y y -intercept of the line 14x−16y=−10 14x - 16y = -10 .\newlineWrite your answer as an integer or as a simplified proper or improper fraction, not as an ordered pair.\newline____
Get tutor helpright-arrow

Posted 1 month ago

Question
Line q q has a slope of −23 \frac{-2}{3} . Line r r is perpendicular to q q . What is the slope of line r r ?\newlineSimplify your answer and write it as a proper fraction, improper fraction, or integer.\newline_____\_\_\_\_\_
Get tutor helpright-arrow

Posted 1 month ago

Question
Find the slope of the line y+3=−118(x+6)y + 3 = -\frac{1}{18}(x + 6).\newlineWrite your answer as an integer or as a simplified proper or improper fraction.
Get tutor helpright-arrow

Posted 1 month ago

Question
Line p p has an equation of y=−8x+6 y = -8x + 6 . Line q q , which is perpendicular to line p p , includes the point (2,−2) (2, -2) . What is the equation of line q q ??\newlineWrite the equation in slope-intercept form. Write the numbers in the equation as simplified proper fractions, improper fractions, or integers.\newline____
Get tutor helpright-arrow

Posted 1 month ago

Question
Is the graph of this equation a horizontal or vertical line?\newlinex=7x = 7\newlineChoices:\newline(A)horizontal line\text{(A)horizontal line}\newline(B)vertical line\text{(B)vertical line}
Get tutor helpright-arrow

Posted 2 months ago

Question
Rewrite the following equation in standard form.\newliney=−10x+7 y = -10x + 7 \newlineHint: The standard form of a linear equation is\newlineAx+By=C Ax + By = C where A A and B B are not both zero, and A A , B B , and C C are integers whose GCF is 11. ______
Get tutor helpright-arrow

Posted 2 months ago

Question
A line with a slope of 99 passes through the point (9,9)(9,9). What is its equation in point-slope form?\newlineUse the specified point in your equation. Write your answer using integers, proper fractions, and improper fractions. Simplify all fractions.\newliney−‾=‾(x−‾)y - \underline{\hspace{1cm}} = \underline{\hspace{1cm}} (x - \underline{\hspace{1cm}})
Get tutor helpright-arrow

Posted 1 month ago

View all (119)