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A line with a slope of −9-9 passes through the points (−4,−8)(-4,-8) and (−6,a)(-6,a). What is the value of aa?\newlinea = ____

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Full solution

Q. A line with a slope of −9-9 passes through the points (−4,−8)(-4,-8) and (−6,a)(-6,a). What is the value of aa?\newlinea = ____
  1. Use Slope Formula: To find the value of ' extit{a}', 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 (−4,−8) (-4, -8) and the slope −9 -9 . We need to find ' extit{a}' for the point (−6,a) (-6, a) .
  2. Denote Points: Let's denote the first point as (x1,y1)=(−4,−8)(x_1, y_1) = (-4, -8) and the second point as (x2,y2)=(−6,a)(x_2, y_2) = (-6, a). The slope of the line is given as −9-9. We can plug these values into the slope formula:\newline(−9)=a−(−8)−6−(−4)(-9) = \frac{a - (-8)}{-6 - (-4)}
  3. Plug Values: Simplify the equation:\newline−9=(a+8)(−6+4)-9 = \frac{(a + 8)}{(-6 + 4)}\newline−9=(a+8)(−2)-9 = \frac{(a + 8)}{(-2)}
  4. Simplify Equation: To find aa, we need to solve for it by multiplying both sides of the equation by −2-2:−9×−2=(a+8)-9 \times -2 = (a + 8)18=a+818 = a + 8
  5. Solve for 'a': Now, subtract 88 from both sides to isolate 'a':\newline18−8=a18 - 8 = a\newline10=a10 = a

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