225k views
4 votes
The point-slope form of the equation of the line that passes through (–5, –1) and (10, –7) is ______. What is the standard form of the equation for this line?

1) 2x – 5y = –15
2) 2x – 5y = –17
3) 2x + 5y = –15
4) 2x + 5y = –17

User Geekfish
by
7.8k points

1 Answer

2 votes

Final answer:

The point-slope form of the equation of the line that passes through (-5, -1) and (10, -7) is y + 1 = (-8/15)(x + 5). Converting this equation to standard form gives us 8x + 15y = -55.

Step-by-step explanation:

To find the equation of the line that passes through the points (-5, -1) and (10, -7), we can use the point-slope form of a linear equation, which is: y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope of the line.

First, let's find the slope of the line using the formula: slope (m) = (y2 - y1) / (x2 - x1). Plugging in the coordinates of the two points, we get: slope (m) = (-7 - (-1)) / (10 - (-5)) = -8/15.

Now we can choose one of the points, such as (-5, -1), and plug it into the point-slope form to find the equation of the line. We have: y - (-1) = (-8/15)(x - (-5)). Simplifying this equation gives us: y + 1 = (-8/15)(x + 5).

Converting this equation to standard form by multiplying both sides by 15 and rearranging the terms, we get: 15y + 15 = -8x - 40. Finally, rearranging the terms to match one of the given options, we have: 8x + 15y = -55. Therefore, the standard form of the equation for this line is option 1) 2x - 5y = -15.

User Syfer
by
7.8k points