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what’s the equation of the line when the line is perpendicular to 7y-x=-35 and passed through the point (-14,8)

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The equation of a line perpendicular to another line can be found by first determining the slope of the given line, and then taking the negative reciprocal of that slope.

To find the slope of the given line 7y - x = -35, we can rearrange the equation into slope-intercept form (y = mx + b), where m represents the slope:

7y - x = -35

Rearranging the equation, we get:

7y = x - 35

Dividing both sides by 7, we have:

y = (1/7)x - 5

So, the slope of the given line is 1/7.

To find the slope of the line perpendicular to this line, we take the negative reciprocal of 1/7.

The negative reciprocal of a number is found by flipping the fraction and changing its sign. So, the negative reciprocal of 1/7 is -7/1 or simply -7.

Now that we have the slope of the line perpendicular to the given line, we can use the point-slope form of a line to find the equation of the line passing through the point (-14,8).

The point-slope form of a line is given by:

y - y1 = m(x - x1)

where (x1, y1) represents the coordinates of a point on the line, and m represents the slope of the line.

Substituting the given values, we have:

y - 8 = -7(x - (-14))

Simplifying the equation, we get:

y - 8 = -7(x + 14)

Expanding the equation, we have:

y - 8 = -7x - 98

Finally, rearranging the equation to slope-intercept form, we have:

y = -7x - 90

Therefore, the equation of the line that is perpendicular to 7y - x = -35 and passes through the point (-14,8) is y = -7x - 90.

User Jim Ferrans
by
8.6k points
5 votes

Answer:

It's y = -7x + 10

User Glen Pierce
by
7.9k points

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