Question

medals!!!!
Write the equation of the line that is perpendicular to the line y =3/2 x + 4 and passes through the point (−6, 3).

Answer 1

slope is \[-\frac{2}{3}\] and point is
\[(-6,3)\] use
\[y-y_1=m(x-x_1)\] with
\[m=-\frac{2}{3}, x_1=-6,y_1=3\]

Answer 2

perpendicular---> slope of the line we want is =-1/k=-1/(3/2)=-2/3
and eq is (y-y1)=k(x-x1)
y-3=-2/3(x+6)
y-3=-2x/3 -4

Answer 3

get what angela wrote

Answer 4

uhh. no, i don't get it... at all!! Can someone explain it to me?

Answer 5

the line tht u r asking must be perpendicular with y =3/2 x + 4
(k=slope=3/2)
so the slope k1 we're looking for is k1=-1/k=-1(3/2)=-2/3
then when u have the slope and a point u use (y-y1)=k(x-x1) this eq...all u do then i replace y1=3 and x1=-6 and u get...
y-3=-2/3(x+6)
y-3=-2x/3 -4

Answer 6

Slope of a perpendicular line is the negative reciprocal of the orginal equation you were given, Hence -2/3
Then you were given two points in which you have to lug in this equation (y-y1) = m (x-x1)
M is the slope which is -2/3and (x1 and y1 are your given coordinates. Plug it into the equation and solve.

Answer 7

okay, i got y= -2/3x+1 is that right? Also, how do you figure out the "m=−23, x1=−6 ,y1=3" part

Answer 8

The solve of a perpendicular line is the negative reciprocal of the orginal equation Your orginal slope of the above equation was 3/2. The negative reciprocal of it is -2/3 Your question asked us to use the coordinates (-6,3) or (x,y). - 6 is x and 3 is y