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A line has a slope of 15\frac{1}{5} and passes through the point (−7,−3)(-7,-3). What is its equation in point-slope form?\newlineUse the specified point in your equation. Write your answer using integers, proper fractions, and improper fractions. Simplify all fractions.\newliney−______=______(x−______)y - \_\_\_\_\_\_ = \_\_\_\_\_\_(x - \_\_\_\_\_\_)

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Q. A line has a slope of 15\frac{1}{5} and passes through the point (−7,−3)(-7,-3). What is its equation in point-slope form?\newlineUse the specified point in your equation. Write your answer using integers, proper fractions, and improper fractions. Simplify all fractions.\newliney−______=______(x−______)y - \_\_\_\_\_\_ = \_\_\_\_\_\_(x - \_\_\_\_\_\_)
  1. Identify Point-Slope Form: Identify the point-slope form of a linear equation.\newlineThe point-slope form of a linear equation is given by y−y1=m(x−x1)y - y_1 = m(x - x_1), where mm is the slope of the line and (x1,y1)(x_1, y_1) is a point on the line.
  2. Substitute Slope and Point: Substitute the given slope and point into the point-slope form.\newlineWe are given the slope m=15m = \frac{1}{5} and the point (−7,−3)(-7, -3). Substituting these values into the point-slope form, we get:\newliney−(−3)=(15)(x−(−7))y - (-3) = \left(\frac{1}{5}\right)(x - (-7))
  3. Simplify Equation: Simplify the equation.\newlineSimplifying the equation, we get:\newliney+3=15(x+7)y + 3 = \frac{1}{5}(x + 7)\newlineThis is the equation of the line in point-slope form.

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