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solong [7]
2 years ago
10

Find an equation for the line that passes through the point P(-5,-3) and is parallel to the line

Mathematics
1 answer:
Aneli [31]2 years ago
4 0

-------------------------------------------------------------------------------------------------------------

Answer:  \textsf{y = -1.75x - 11.75}

-------------------------------------------------------------------------------------------------------------

Given:  \textsf{Goes through (-5, -3) and parallel to 7x + 4y = 10}

Find:  \textsf{The equation in slope-intercept form}

Solution: We need to first solve for y in the equation that was provided so we can determine the slope.  Then we plug in the values into the point-slope form, distribute, simplify, and solve for y to get our final equation.

<u>Subtract 7x from both sides</u>

  • \textsf{7x - 7x + 4y = 10 - 7x}
  • \textsf{4y = 10 - 7x}

<u>Divide both sides by 4</u>

  • \textsf{4y/4 = (10 - 7x)/4}
  • \textsf{y = (10 - 7x)/4}
  • \textsf{y = 10/4 - 7x/4}
  • \textsf{y = 2.5 - 1.75x}

<u>Plug in the values</u>

  • \textsf{y - y}_1\textsf{ = m(x - x}_1\textsf{)}
  • \textsf{y - (-3) = -1.75(x - (-5))}

<u>Simplify and distribute</u>

  • \textsf{y + 3 = -1.75(x + 5)}
  • \textsf{y + 3 = (-1.75 * x) + (-1.75 * 5)}
  • \textsf{y + 3 = -1.75x - 8.75}

<u>Subtract 3 from both sides</u>

  • \textsf{y + 3 - 3 = -1.75x - 8.75 - 3}
  • \textsf{y = -1.75x - 8.75 - 3}
  • \textsf{y = -1.75x - 11.75}

Therefore, the final equation in slope-intercept form that follows the information that was provided is y = -1.75x - 11.75

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Step-by-step explanation:

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How do i find LM? On number 2 .
Masja [62]
Given: KL is a tangent to the circle.
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</em>
Thus, we can say KL = LM, as they lie on a common circle, and are tangents to such circle.

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