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A line that includes the points (−4,h)(-4,h) and (−2,−2)(-2,-2) has a slope of −6-6. What is the value of hh?\newlineh = ____

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Q. A line that includes the points (−4,h)(-4,h) and (−2,−2)(-2,-2) has a slope of −6-6. What is the value of hh?\newlineh = ____
  1. Slope Formula Application: The slope of a line passing through two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is given by the formula:\newlineslope = y2−y1x2−x1\frac{y_2 - y_1}{x_2 - x_1}\newlineWe are given the slope of the line as −6-6, and the coordinates of the two points as (−4,h)(-4, h) and (−2,−2)(-2, -2). We can plug these values into the slope formula to find hh.
  2. Solving for hh: Using the slope formula with our given values:\newline−6=(−2−h)/(−2−(−4))-6 = (-2 - h) / (-2 - (-4))\newlineSimplify the denominator:\newline−6=(−2−h)/(2)-6 = (-2 - h) / (2)
  3. Isolating hh: To find the value of hh, we need to solve for hh in the equation:\newline−6=(−2−h)/2-6 = (-2 - h) / 2\newlineMultiply both sides by 22 to eliminate the denominator:\newline−6×2=−2−h-6 \times 2 = -2 - h\newline−12=−2−h-12 = -2 - h
  4. Final Solution: Now, we add 22 to both sides of the equation to isolate hh: \newline−12+2=−2+2−h-12 + 2 = -2 + 2 - h\newline−10=−h-10 = -h
  5. Final Solution: Now, we add 22 to both sides of the equation to isolate hh: \newline−12+2=−2+2−h-12 + 2 = -2 + 2 - h\newline−10=−h-10 = -hFinally, we multiply both sides by −1-1 to solve for hh: \newline−10×−1=−h×−1-10 \times -1 = -h \times -1\newline10=h10 = h

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