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The points (−1,−8)(-1,-8) and (−3,j)(-3,j) fall on a line with a slope of −8-8. What is the value of jj?\newlinej=___j = \_\_\_

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Q. The points (−1,−8)(-1,-8) and (−3,j)(-3,j) fall on a line with a slope of −8-8. What is the value of jj?\newlinej=___j = \_\_\_
  1. Use Slope Formula: To find the value of jj, we can use the slope formula, which is y2−y1x2−x1=slope\frac{y_2 - y_1}{x_2 - x_1} = \text{slope}, where (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) are the coordinates of two points on the line.
  2. Plug in Values: We know the slope of the line is −8-8, and we have the coordinates of one point (−1,−8)(-1, -8). We can plug these values into the slope formula along with the xx-coordinate of the second point, which is −3-3. Let's denote the yy-coordinate of the second point as jj.
  3. Simplify Equation: The slope formula with our known values is (−8−j)/(−1−(−3))=−8(-8 - j) / (-1 - (-3)) = -8. Simplifying the denominator, we get (−8−j)/(2)=−8(-8 - j) / (2) = -8.
  4. Isolate jj: To find the value of jj, we need to solve the equation (−8−j)/2=−8(-8 - j) / 2 = -8. Multiplying both sides by 22 to isolate the term with jj gives us −8−j=−16-8 - j = -16.
  5. Solve for j: Now, we add 88 to both sides of the equation to solve for jj: −j=−16+8-j = -16 + 8, which simplifies to −j=−8-j = -8.
  6. Final Value of j: Finally, we multiply both sides by −1-1 to find the value of jj: j=8j = 8.

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