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The equation of line 
k is 
y+10= 
3(x+3). Perpendicular to line 
k is line 
ℓ, which passes through the point 
(5,-5). What is the equation of line 
ℓ ?
Write the equation in slope-intercept form. Write the numbers in the equation as simplified proper fractions, improper fractions, or integers.

The equation of line k k is y+10= y+10= 3(x+3) 3(x+3) . Perpendicular to line k k is line \ell , which passes through the point (5,5) (5,-5) . What is the equation of line \ell ?\newlineWrite the equation in slope-intercept form. Write the numbers in the equation as simplified proper fractions, improper fractions, or integers.

Full solution

Q. The equation of line k k is y+10= y+10= 3(x+3) 3(x+3) . Perpendicular to line k k is line \ell , which passes through the point (5,5) (5,-5) . What is the equation of line \ell ?\newlineWrite the equation in slope-intercept form. Write the numbers in the equation as simplified proper fractions, improper fractions, or integers.
  1. Determine Slope of Line k: Determine the slope of line k.\newlineThe equation of line k is given in point-slope form: y+10=3(x+3)y + 10 = 3(x + 3).\newlineTo find the slope, we need to rewrite it in slope-intercept form (y=mx+by = mx + b), where mm is the slope.\newliney+10=3x+9y + 10 = 3x + 9\newlineSubtract 1010 from both sides to isolate yy.\newliney=3x+910y = 3x + 9 - 10\newliney=3x1y = 3x - 1\newlineThe slope of line k is 33.
  2. Find Slope of Line \ell: Find the slope of line \ell, which is perpendicular to line kk. Slopes of perpendicular lines are opposite reciprocals. The slope of line kk is 33, so the slope of line \ell will be the negative reciprocal of 33. The negative reciprocal of 33 is 13-\frac{1}{3}. Therefore, the slope of line \ell is 13-\frac{1}{3}.
  3. Use Point and Slope to Find Equation: Use the point (5,5)(5, -5) and the slope 13-\frac{1}{3} to find the equation of line \ell. We will use the point-slope form of the equation, which is yy1=m(xx1)y - y_1 = m(x - x_1), where (x1,y1)(x_1, y_1) is a point on the line and mm is the slope. Plugging in the point (5,5)(5, -5) and the slope 13-\frac{1}{3}, we get: y(5)=13(x5)y - (-5) = -\frac{1}{3}(x - 5) y+5=13x+53y + 5 = -\frac{1}{3}x + \frac{5}{3}
  4. Solve for y in Slope-Intercept Form: Solve for y to put the equation in slope-intercept form.\newlineSubtract 55 from both sides to isolate yy.\newliney=13x+535y = -\frac{1}{3}x + \frac{5}{3} - 5\newlineTo combine the terms, we need a common denominator, which is 33.\newliney=13x+53153y = -\frac{1}{3}x + \frac{5}{3} - \frac{15}{3}\newliney=13x103y = -\frac{1}{3}x - \frac{10}{3}\newlineNow we have the equation of line \ell in slope-intercept form.

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