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Line t is expressed with the equation y=-5/3 +3 what is the equation of the line that is parallel to line t through the point (-1,7)?

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Final answer:

The equation of the line parallel to line t with equation y = -5/3x + 3 and passing through the point (-1,7) is y = (-5/3)x + 16/3.

Step-by-step explanation:

The question asks about finding the equation of a line that is parallel to another line and goes through a specific point. Given the equation y = -5/3 + 3, it seems there may be a typo, and it's assumed to be y = (-5/3)x + 3 to have both a slope and a y-intercept. For parallel lines, the slopes must be identical. The slope here is -5/3. The new line will thus have the same slope and pass through point (-1,7).

To find the equation of the new line, we use the point-slope form:

y - y1 = m(x - x1)

y - 7 = (-5/3)(x - (-1))

y - 7 = (-5/3)(x + 1)

Now we put this into slope-intercept form (y = mx + b) by distributing the slope and moving the 7 over:

y = (-5/3)x - (5/3) + 7

y = (-5/3)x + (21/3 - 5/3)

y = (-5/3)x + 16/3

This is the equation of the line parallel to line t that passes through the point (-1,7).

User Stevehayter
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