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

2 Answers
Mar 6, 2018

Let,the equation of line be y=mx+cy=mx+c, where, mm is its slope and cc is the YY intercept.

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

Given, y+5=3(x-2)=y=3x-11y+5=3(x2)=y=3x11

So,its slope is 33

Now,for two lines to be mutually perpendicular,their product of slope must be -11

So, m*3=-1m3=1

or. m=-1/3m=13

So,our required line equation becomes, y=-1/3 x+cy=13x+c

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

So, 2=(-1/3)*6 +c2=(13)6+c

or, c=4c=4

So,the equation of the line is y=-1/3 x+4y=13x+4

or, 3y+x=123y+x=12 graph{3y+x=12 [-10, 10, -5, 5]}

Mar 6, 2018

y-2 = -1/3(x-6)y2=13(x6)

Or y = -1/3x +4y=13x+4

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

Or x+3y-12=0x+3y12=0

Explanation:

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

y = mx+c," and " ax +by +c =0" "y=mx+c, and ax+by+c=0 are well known.

Another is y-y_1 = m(x-x_1)yy1=m(xx1)

where (x_1, y_1)(x1,y1) is a point and mm 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