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If the line passing through the points (1,a) and (4,−3) is parallel to the line passing through the points (2,0) and (−4, a +9) whint is the value of a?

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Answer:

a = 3

Explanation:

Relationship between the slopes of parallel lines:

  • The slopes of parallel lines are the same.

Using the slope formula to find the slope:

Given two points on a line, we can find the slope of the line using the slope formula, which is given by:

m = (y2 - y1) / (x2 - x1), where

  • m is the slope,
  • (x1, y1) is one point on the line,
  • and (x2, y2) is another point.

Finding the value of a that makes the lines parallel:

Since we need to find the value of a that makes the slopes the same, we set the slope formulas equal to each other and solve for a.

This means that:

  • on the left-hand side, we substitute (1, a) and (4, -3) for (x1, y1) and (x2, y2) in the slope formula,
  • and on the right-hand side, we substitute (2, 0) and (-4, a + 9) for (x1, y1) and (x2, y2) in the slope formula.

(-3 - a) / (4 - 1) = (a + 9 - 0) / (-4 - 2)

(-3 - a) / (3) = (a + 9) / (-6)

  • Note that dividing by a number is the same as multiplying by its reciprocal, meaning that x / 3 = 1/3x and x / 4 = 1/4x.

1/3(-3 - a) = -1/6(a + 9)

(-1 - 1/3a = -1/6a - 3/2) + 1

(-1/3a = -1/6a - 1/2) + 1/6a

(-1/6a = -1/2) / (-1/6)

(-1/6a = -1/2) * (-6)

a = 3

Thus, a being 3 makes the liens parallel.

Checking the validity of the answer:

  • Now we can check that our answer is correct by finding the slopes of the two lines with the slope formula and substituting 3 for a.
  • This means that our points for the first line are (1, 3) and (4, -3), while our points for the other line are (2, 0) and (-4, 12).

If we get the same slope for both lines, we've correctly found the correct value of a:

Finding the slope of (1, 3) and (4, -3):

m = (-3 - 3) / (4 - 1)

m = (-6) / (3)

m = -2

Finding the slope of (2, 0) and (-4, 12):

m = (12 - 0) / (-4 - 2)

m = (12) / (-6)

m = -2

Thus, we've correctly determined the value of a that makes the two lines parallel.

User Casey Watson
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