What is the equation of the line in slope-intercept that is perpendicular to the line #4y - 2 = 3x# and passes through the point (6,1)?

Question

Wyatt Anderson · Accepted Answer

Let,the equation of the line required is #y=mx+c# where, #m# is the slope and #c# is the #Y# intercept. Given equation of line is #4y-2=3x# or, #y=3/4 x +1/2# Now,for these two lines to be perpendicular product of their slope has to be #-1# i.e #m(3/4)=-1# so, #m=-4/3# Hence,the equation becomes, #y=-4/3x+c# Given,that this line passes through #(6,1)#,putting the values in our equation we get, #1=(-4/3)*6 +c# or, #c=9# So,the required equation becomes, #y=-4/3 x+9# or, #3y+4x=27# graph{3y+4x=27 [-10, 10, -5, 5]}

Allison Allen · Answer

undefined To find the equation of the line perpendicular to the given line $4y - 2 = 3x$ in slope-intercept form and passing through the point $(6,1)$, follow these steps:

1. First, rearrange the given equation into slope-intercept form $y = mx + b$, where $m$ is the slope and $b$ is the y-intercept.
   $4y - 2 = 3x$ becomes $y = \frac{3}{4}x + \frac{1}{2}$.
   
2. Determine the slope of the given line. The slope of the line $y = \frac{3}{4}x + \frac{1}{2}$ is $m = \frac{3}{4}$.

3. The slope of a line perpendicular to this line will have a negative reciprocal slope. So, the slope of the perpendicular line is $-\frac{4}{3}$.

4. Use the point-slope form of a linear equation: $y - y_1 = m(x - x_1)$, where $(x_1, y_1)$ is the given point and $m$ is the slope.
   Plug in the point $(6,1)$ and the slope $-\frac{4}{3}$.
   $y - 1 = -\frac{4}{3}(x - 6)$.

5. Simplify the equation to slope-intercept form by distributing and rearranging terms.
   $y - 1 = -\frac{4}{3}x + 8$.
   $y = -\frac{4}{3}x + 9$.

So, the equation of the line in slope-intercept form that is perpendicular to the line $4y - 2 = 3x$ and passes through the point $(6,1)$ is $y = -\frac{4}{3}x + 9$.