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A line has a slope of 33 and includes the points (−5,j)(-5,j) and (−7,−4)(-7,-4). What is the value of jj?\newlinej=___j = \_\_\_

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Full solution

Q. A line has a slope of 33 and includes the points (−5,j)(-5,j) and (−7,−4)(-7,-4). 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 33, and we have the coordinates of one point (−7,−4)(-7, -4). We can plug these values into the slope formula along with the coordinates of the second point (−5,j)(-5, j) to find jj.
  3. Simplify Equation: Using the slope formula: (j−(−4))/(−5−(−7))=3(j - (-4)) / (-5 - (-7)) = 3. This simplifies to (j+4)/2=3(j + 4) / 2 = 3.
  4. Eliminate Denominator: To solve for jj, we multiply both sides of the equation by 22 to get rid of the denominator: 2×(j+42)=2×32 \times \left(\frac{j + 4}{2}\right) = 2 \times 3.
  5. Isolate Variable: This simplifies to j+4=6j + 4 = 6. Now, we just need to subtract 44 from both sides to solve for jj.
  6. Final Calculation: Subtracting 44 from both sides gives us j=6−4j = 6 - 4.
  7. Find Value of j: Calculating the difference, we find that j=2j = 2.

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