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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.

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Q. 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.
  1. Find Slope of Line pp: Line pp has a slope of 8-8, as we can see from its equation y=8x+6y = -8x + 6. Since line qq is perpendicular to line pp, we need to find the slope of line qq, which will be the negative reciprocal of the slope of line pp.
  2. Determine Slope of Line qq: The negative reciprocal of 8-8 is 18\frac{1}{8}. Therefore, the slope of line qq is 18\frac{1}{8}.
  3. Calculate Y-Intercept of Line q: Now we need to find the y-intercept of line q. We know that line q passes through the point (2,2)(2, -2) and has a slope of 18\frac{1}{8}. We can use the point-slope form of the equation of a line to find the y-intercept. The point-slope form is yy1=m(xx1)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.
  4. Apply Point-Slope Form: Plugging in the slope and the point (2,2)(2, -2) into the point-slope form, we get y(2)=18(x2)y - (-2) = \frac{1}{8}(x - 2). Simplifying this, we get y+2=18x14y + 2 = \frac{1}{8}x - \frac{1}{4}.
  5. Isolate y in Equation: To get the equation in slope-intercept form, we need to isolate y. Subtracting 22 from both sides of the equation, we get y=18x142y = \frac{1}{8}x - \frac{1}{4} - 2.
  6. Simplify Final Equation: Simplifying the equation further, we combine the constant terms 14-\frac{1}{4} and 2-2. Since 2-2 is the same as 84-\frac{8}{4}, we get y=18x1484y = \frac{1}{8}x - \frac{1}{4} - \frac{8}{4}, which simplifies to y=18x94y = \frac{1}{8}x - \frac{9}{4}.

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