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A line has a slope of 3-3 and includes the points (6,f)(-6,f) and (4,1)(-4,1). What is the value of ff?\newlinef = ____

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Q. A line has a slope of 3-3 and includes the points (6,f)(-6,f) and (4,1)(-4,1). What is the value of ff?\newlinef = ____
  1. Given Slope and Points: We are given the slope of the line and two points that lie on the line. The slope formula is (y2y1)/(x2x1)(y_2 - y_1) / (x_2 - x_1), where (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) are the coordinates of the two points on the line. We can use this formula to find the value of ff.
  2. Plug in Values: Let's plug in the values we know into the slope formula. We have the slope m=3m = -3, point 11 (x1,y1)=(6,f)(x_1, y_1) = (-6, f), and point 22 (x2,y2)=(4,1)(x_2, y_2) = (-4, 1). The formula becomes:\newline3=1f4(6)-3 = \frac{1 - f}{-4 - (-6)}
  3. Simplify Denominator: Simplify the denominator of the fraction on the right side of the equation:\newline3=1f4+6-3 = \frac{1 - f}{-4 + 6}
  4. Multiply by 22: Now we have:\newline3=1f2-3 = \frac{1 - f}{2}\newlineTo find the value of f, we need to solve for f. Let's multiply both sides of the equation by 22 to get rid of the denominator:\newline3×2=(1f)×22-3 \times 2 = (1 - f) \times \frac{2}{2}
  5. Isolate f: Simplifying both sides of the equation gives us:\newline6=1f-6 = 1 - f\newlineNow, we need to isolate f. We can do this by subtracting 11 from both sides of the equation:\newline61=11f-6 - 1 = 1 - 1 - f
  6. Multiply by 1-1: This simplifies to:\newline7=f-7 = -f\newlineTo solve for ff, we multiply both sides by 1-1:\newline1×(7)=1×(f)-1 \times (-7) = -1 \times (-f)
  7. Final Value of f: Finally, we have:\newlinef = 77\newlineWe have found the value of ff.

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