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The points (−1,−10)(-1,-10) and (−3,p)(-3,p) fall on a line with a slope of −3-3. What is the value of pp?\newlinep=___p = \_\_\_

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Q. The points (−1,−10)(-1,-10) and (−3,p)(-3,p) fall on a line with a slope of −3-3. What is the value of pp?\newlinep=___p = \_\_\_
  1. Use Slope Formula: To find the value of pp, 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 −3-3, and we have the coordinates of one point (−1,−10)(-1, -10) and the xx-coordinate of the second point (−3)(-3). We can plug these values into the slope formula to find pp (y2y_2).
  3. Simplify Equation: Using the slope formula with our known values, we get (−10−p)/(−1−(−3))=−3(-10 - p) / (-1 - (-3)) = -3. Simplifying the denominator, we have (−10−p)/(2)=−3(-10 - p) / (2) = -3.
  4. Multiply by 22: To find the value of pp, we need to solve the equation (−10−p)/2=−3(-10 - p) / 2 = -3. We can start by multiplying both sides of the equation by 22 to get rid of the denominator.
  5. Add 1010: Multiplying both sides by 22 gives us −10−p=−3×2-10 - p = -3 \times 2, which simplifies to −10−p=−6-10 - p = -6.
  6. Isolate pp: Now, we can add 1010 to both sides of the equation to isolate pp on one side. This gives us −p=−6+10-p = -6 + 10.
  7. Solve for pp: Solving for pp, we get −p=4-p = 4. To get pp, we multiply both sides by −1-1, which gives us p=−4p = -4.

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