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A line has a slope of 1010 and includes the points (0,p)(0,p) and (−1,−5)(-1,-5). What is the value of pp?\newlinep = ____

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Full solution

Q. A line has a slope of 1010 and includes the points (0,p)(0,p) and (−1,−5)(-1,-5). What is the value of pp?\newlinep = ____
  1. Identify Points and Slope: Identify the given points and the slope.\newlinePoints: (0,p)(0, p) and (−1,−5)(-1, -5)\newlineSlope: 1010\newlineUse the slope formula to set up the equation.\newlineSlope (m)=y2−y1x2−x1(m) = \frac{y_2 - y_1}{x_2 - x_1}\newline10=p−(−5)0−(−1)10 = \frac{p - (-5)}{0 - (-1)}
  2. Set Up Equation: Simplify the equation.\newline10=p+5110 = \frac{p + 5}{1}\newlineMultiply both sides by 11 to isolate (p+5)(p + 5).\newline10×1=(p+5)10 \times 1 = (p + 5)
  3. Simplify Equation: Solve for pp.10=p+510 = p + 5Subtract 55 from both sides to find the value of pp.10−5=p10 - 5 = pp=5p = 5

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