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A line with a slope of 11 passes through the points (9,−6)(9,-6) and (7,t)(7,t). What is the value of tt?\newlinet = ____

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Q. A line with a slope of 11 passes through the points (9,−6)(9,-6) and (7,t)(7,t). What is the value of tt?\newlinet = ____
  1. Use Slope Formula: To find the value of tt, we can use the slope formula, which is y2−y1x2−x1=slope\frac{y_2 - y_1}{x_2 - x_1} = \text{slope}. Here, we have one point (9,−6)(9, -6) and the slope of 11. We need to find the y-coordinate (tt) of the second point (7,t)(7, t).
  2. Plug in Values: Let's plug in the values we have into the slope formula. We'll use (9,−6)(9, -6) as (x1,y1)(x_1, y_1) and (7,t)(7, t) as (x2,y2)(x_2, y_2). The slope is given as 11. So, we have: t−(−6)7−9=1\frac{t - (-6)}{7 - 9} = 1
  3. Simplify Equation: Simplify the equation: (t+6)/(7−9)=1(t + 6) / (7 - 9) = 1
  4. Calculate Denominator: Calculate the denominator:\newline7−9=−27 - 9 = -2\newlineSo, (t+6)/−2=1(t + 6) / -2 = 1
  5. Isolate t: To find t, we need to solve for t in the equation (t+6)/−2=1(t + 6) / -2 = 1. Multiply both sides by −2-2 to isolate t:\newline(t+6)=−2×1(t + 6) = -2 \times 1
  6. Perform Multiplication: Now, perform the multiplication: (t+6)=−2(t + 6) = -2
  7. Solve for t: Finally, subtract 66 from both sides to solve for tt: \newlinet=−2−6t = -2 - 6\newlinet=−8t = -8

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