Question

Find the equation of a line perpendicular to y=5x-3 and which passes through (-3,4)

Answer 1

To find the equation of a line perpendicular to y=5x-3 and which passes through (-3,4), we first need to find the slope of the line y=5x-3. The slope of this line is 5.

Since we want to find the equation of a line that is perpendicular to y=5x-3, we know that the slope of this new line will be the negative reciprocal of 5. The negative reciprocal of 5 is -1/5.

Now that we know the slope of the new line, we can use the point-slope form of a line to find its equation. The point-slope form of a line is y-y1=m(x-x1), where m is the slope of the line and (x1,y1) is a point on the line.

Using the point (-3,4) and the slope -1/5, we get:

y-4=-1/5(x-(-3)

y-4=-1/5(x+3)

y-4=-1/5x-3/5

y=-1/5x+23/5

Therefore, the equation of a line perpendicular to y=5x-3 and which passes through (-3,4) is y=-1/5x+23/5.

Answer 2

Answer: Hope this helps :)

Explanation:

The slope of the given line y = 5x - 3 is 5. The slope of a line perpendicular to it would be the negative reciprocal of 5, which is -1/5. Using the point-slope form of a line, we can find the equation of the line perpendicular to y = 5x - 3 that passes through the point (-3,4) as follows: y - y1 = m(x - x1), where (x1,y1) is the point (-3,4) and m is the slope -1/5. Substituting these values into the equation gives us: y - 4 = (-1/5)(x + 3). Simplifying this equation gives us: y = (-1/5)x - 3/5 + 4, or y = (-1/5)x + 17/5.