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The points (−6,−2)(-6,-2) and (−5,j)(-5,j) fall on a line with a slope of −6-6. What is the value of jj?\newlinej=___j = \_\_\_

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Q. The points (−6,−2)(-6,-2) and (−5,j)(-5,j) fall on a line with a slope of −6-6. What is the value of jj?\newlinej=___j = \_\_\_
  1. Given Points and Slope: We are given two points on a line: (−6,−2)(-6,-2) and (−5,j)(-5,j). We also know the slope of the line is −6-6. The slope formula is (y2−y1)/(x2−x1)(y_2 - y_1) / (x_2 - x_1). Let's use the coordinates of the two points to find the value of jj.
  2. Plug in Values: First, let's plug in the values we know into the slope formula: (−6)=(j−(−2))/(−5−(−6))(-6) = (j - (-2)) / (-5 - (-6)).
  3. Simplify Equation: Simplify the equation: (−6)=(j+2)1(-6) = \frac{(j + 2)}{1}.
  4. Multiply to Eliminate Denominator: Multiply both sides of the equation by 11 to get rid of the denominator: −6=j+2-6 = j + 2.
  5. Subtract to Solve for j: Subtract 22 from both sides to solve for jj: −6−2=j-6 - 2 = j.
  6. Calculate j Value: Calculate the value of j: j=−8j = -8.

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