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Complete the equation of the line through (-10,3) and (-8,-8). Use exact numbers. y=

Complete the equation of the line through (−10,3) (-10,3) and (−8,−8) (-8,-8) . Use exact numbers.\newliney= y=

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Q. Complete the equation of the line through (−10,3) (-10,3) and (−8,−8) (-8,-8) . Use exact numbers.\newliney= y=
  1. Calculate the slope: To find the equation of a line, we need to determine the slope mm of the line using the formula m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}, where (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) are the coordinates of the two points the line passes through.\newlineLet's calculate the slope using the points (−10,3)(-10, 3) and (−8,−8)(-8, -8).\newlinem=−8−3−8−(−10)m = \frac{-8 - 3}{-8 - (-10)}\newlinem=−112m = \frac{-11}{2}
  2. Use point-slope form: Now that we have the slope m=−112m = -\frac{11}{2}, we can use the point-slope form of the equation of a line, which is y−y1=m(x−x1)y - y_1 = m(x - x_1), where (x1,y1)(x_1, y_1) is one of the points the line passes through. We can use either point, but let's use (−10,3)(-10, 3).\newliney−3=(−112)(x−(−10))y - 3 = \left(-\frac{11}{2}\right)(x - (-10))\newliney−3=(−112)(x+10)y - 3 = \left(-\frac{11}{2}\right)(x + 10)
  3. Distribute the slope: Next, we distribute the slope −112-\frac{11}{2} across the (x+10)(x + 10).\newliney - 33 = \left(-\frac{1111}{22}\right)x - \left(\frac{1111}{22}\right)\cdot 1010\newliney - 33 = \left(-\frac{1111}{22}\right)x - 5555
  4. Solve for y: Finally, we add 33 to both sides of the equation to solve for yy.\newliney=(−112)x−55+3y = \left(-\frac{11}{2}\right)x - 55 + 3\newliney=(−112)x−52y = \left(-\frac{11}{2}\right)x - 52

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