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A line that includes the points (9,−4)(9,-4) and (10,j)(10,j) has a slope of 55. What is the value of jj?\newlinej=___j = \_\_\_

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Full solution

Q. A line that includes the points (9,−4)(9,-4) and (10,j)(10,j) has a slope of 55. What is the value of jj?\newlinej=___j = \_\_\_
  1. Identify Given Points and Slope: Identify the given points and the slope.\newlineWe have the points (9,−4)(9, -4) and (10,j)(10, j) and the slope of the line is 55.\newlineWe will use the slope formula which is (y2−y1)/(x2−x1)=slope.(y_2 - y_1) / (x_2 - x_1) = \text{slope}.
  2. Plug Values into Slope Formula: Plug the values into the slope formula.\newlineUsing the points (9,−4)(9, -4) as (x1,y1)(x_1, y_1) and (10,j)(10, j) as (x2,y2)(x_2, y_2), we get:\newline5=j−(−4)10−95 = \frac{j - (-4)}{10 - 9}\newline5=j+415 = \frac{j + 4}{1}
  3. Simplify Equation and Solve: Simplify the equation and solve for jj. Since the denominator is 11, we can multiply both sides by 11 to get rid of the fraction: 5×1=(j+4)×15 \times 1 = (j + 4) \times 1 5=j+45 = j + 4
  4. Isolate Variable jj: Isolate jj by subtracting 44 from both sides of the equation.\newline5−4=j+4−45 - 4 = j + 4 - 4\newline1=j1 = j

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