Question

Write the equation of a line perpendicular to y=-1/3x-10 that passes through (-1,5)

Answer 1

Problem:

Write the equation of a line perpendicular to y=-1/3x-10 that passes through (-1,5).

Solution:

By definition, the slope-intercept equation of a line is given by the following formula:

`y\text{ = mx+b}`

where m is the slope of the line and b is the y-coordinate of the y-intercept. Now, if we have the equation of a line, given by the formula:

`y\text{ = -}(1)/(3)x\text{ -10}`

then, the slope of this line is -1/3. Now, perpendicular lines have slopes that are the opposite of the reciprocal of each other. In this case, if the slope of the first line is -1/3, the reciprocal of -1/3 is -3, so the opposite of the reciprocal is, therefore 3. Then, the slope of the line perpendicular to y=-1/3x-10 is m = 3, and its equation would be:

y = -3x +b

now, to find b, we take any point of the above line and replace it in the above equation. Take for example, (x,y) = (-1,5), then we have:

5 = -3(-1) + b

this is equivalent to:

5 = 3 + b

solving for b, we get:

b = 2.

then, we can conclude that the equation of a line perpendicular to y=-1/3x-10 that passes through (-1,5) is

`y\text{ = 3x + 2}`