How do you find the equation of the line perpendicular to y+5=3(x-2) that passes through the point (6, 2)?

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Answer 1 · 2018-03-06T18:16:42

Let,the equation of line be $y = m x + c$ $y=mx+c$ , where, $m$ $m$ is its slope and $c$ $c$ is the $Y$ $Y$ intercept.

Now,arranging the given equation in the above mentioned form to get its slope,

Given, $y + 5 = 3 (x - 2) = y = 3 x - 11$ $y+5=3(x-2)=y=3x-11$

So,its slope is $3$ $3$

Now,for two lines to be mutually perpendicular,their product of slope must be $- 1$ $-1$

So, $m \cdot 3 = - 1$ $m*3=-1$

or. $m = - \frac{1}{3}$ $m=-1/3$

So,our required line equation becomes, $y = - \frac{1}{3} x + c$ $y=-1/3 x+c$

Now,given,that the line passes through $(6, 2)$ $(6,2)$ ,so putting the values in the equation to get $c$ $c$

So, $2 = (- \frac{1}{3}) \cdot 6 + c$ $2=(-1/3)*6 +c$

or, $c = 4$ $c=4$

So,the equation of the line is $y = - \frac{1}{3} x + 4$ $y=-1/3 x+4$

or, $3 y + x = 12$ $3y+x=12$ graph{3y+x=12 [-10, 10, -5, 5]}

Answer 2 · 2018-03-06T19:04:38

$y - 2 = - \frac{1}{3} (x - 6)$ $y-2 = -1/3(x-6)$

Or $y = - \frac{1}{3} x + 4$ $y = -1/3x +4$

Or $x + 3 y = 12$ $x+3y =12$

Or $x + 3 y - 12 = 0$ $x+3y-12=0$

There are several forms for the equation of a straight line:

$y = m x + c, and a x + b y + c = 0$ $y = mx+c," and " ax +by +c =0" "$ are well known.

Another is $y - y_{1} = m (x - x_{1})$ $y-y_1 = m(x-x_1)$

where $(x_{1}, y_{1})$ $(x_1, y_1)$ is a point and $m$ $m$ is the slope.

This is exactly the form we have for the given equation:

$y+5=color(blue)(3)(x-2)" "rarr :. color(blue)(m= 3)$

If lines are perpendicular, the product of their slopes is $-1$

One is the negative reciprocal of the other, so:

$m_1 = 3 " "rArr" "m_2 =-1/3$

Using $m= -1/3$ and the point $(6,2)$

$y-y_1 = m(x-x_1)" "$ gives $" "y-2 = -1/3(x-6)$

Or in another form:

$y= -1/3x+2+2" "rarr y = -1/3x +4$

Or even

$3y =-x+12" "rarr" "x+3y =12$