How do you write an equation of a line passing through (-3, 4), perpendicular to #3y=x-2#?

Question

How do you write an equation of a line passing through (-3, 4),  perpendicular to #3y=x-2#?

Julia Adams · Accepted Answer

#3x+y+5=0# is the required equation of the straight line. graph{(3x+y+5)(x-3y-2)=0 [-8.44, 2.66, -4.17, 1.38]}

Any line perpendicular to #ax+by+c=0# is #bx-ay+k=0# where k is constant. Given equation is #rarr3y=x-2# #rarrx-3y=2# Any line perpendicular to #x-3y=2# will be #3x+y+k=0# As #3x+y+k=0# passes through #(-3,4)#, we have, #rarr3*(-3)+4+k=0# #rarr-9+4+k=0# #rarrk=5# So, the required equation of the straight line is #3x+y+5=0#

Nathan Axel · Answer

undefined To write the equation of a line passing through a given point and perpendicular to another line, follow these steps:

1. Determine the slope of the given line. In this case, the given line is $3y = x - 2$. Rewrite the equation in slope-intercept form, $y = mx + b$, where $m$ is the slope. The slope of the given line is $1/3$.

2. The slope of a line perpendicular to another line is the negative reciprocal of the slope of the given line. So, the slope of the perpendicular line is $-1/m$.

3. Use the point-slope form of a linear equation, $y - y_1 = m(x - x_1)$, where $m$ is the slope and $(x_1, y_1)$ is the given point.

4. Substitute the values of the given point $(-3, 4)$ and the slope $-1/m$ into the point-slope form.

5. Simplify the equation to obtain the equation of the perpendicular line.

Given the steps above, the equation of the line passing through $(-3, 4)$ and perpendicular to $3y = x - 2$ is $y - 4 = -3(x + 3)$.