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A line that includes the points (−3,j)(-3,j) and (0,3)(0,3) has a slope of 22. What is the value of jj?\newlinej=___j = \_\_\_

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Full solution

Q. A line that includes the points (−3,j)(-3,j) and (0,3)(0,3) has a slope of 22. What is the value of jj?\newlinej=___j = \_\_\_
  1. Use Slope Formula: To find jj, we need to use the slope formula, which is (y2−y1)/(x2−x1)(y_2 - y_1) / (x_2 - x_1). Here, we know the slope (mm) is 22, and the points are (−3,j)(-3, j) and (0,3)(0, 3).
  2. Apply Slope Formula: Plugging in the values, the slope formula becomes (3−j)/(0−(−3))=2(3 - j) / (0 - (-3)) = 2. Simplifying the denominator, we get (3−j)/3=2(3 - j) / 3 = 2.
  3. Isolate jj: To isolate jj, first multiply both sides by 33: 3−j=63 - j = 6.
  4. Solve for j: Now, solve for j: 3−6=j3 - 6 = j, which simplifies to j=−3j = -3.

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