Question

The equation for line j can be written as y = 2x + 8. Another line k is perpendicular
to line j and passes through the point (6, -6). Choose the equation for line k.
Slope Intercept: y = mx + b
Point slope: (Y - Y1 ) = m (X - X1 )
A y =
1
2 – 3
2
1
y = - 2 - 3
2
B
y = -2x – 3
3
y=- 2 + 3
2

Answer 1

Answer:

Solution given:

equation of line j is:

y=2x+8

comparing above equation with y=mx+c

we get

m=2

let slope of another line k be M

since lines are perpendicular

their product is -1

so

m *M=-1

2M=-1

M=-½

since it passes from (6,-6)

we have

(y-y1)=M(x-x1)

(y+6)=-½(x-6)

2y+12=-x+6

x+2y+12-6=0

x+2y+6=0 is a required equation of line k.

Answer 2

Answer:

$\displaystyle y = - \frac{1}{2} x - 3$

Step-by-step explanation:

we are given that,

$\displaystyle E _{ j}: y = 2x + 8$

and it's perpendicular to $E_k$

since $E_k$ is perpendicular to $E_j$ the slope of the equation of k has to be -½

because we know that

$\displaystyle m_{ \text{perpendicular}} = - \frac{1}{m}$

we are also given a point where the perpendicular line passes as we got the slope and a point we can consider using point-slope form of linear equation to figure out the perpendicular line

remember the point slope form

$\displaystyle y - y_{1} = m(x - x_{1})$

we got that, y1=-6,x1=6 and m=-½ thus,

substitute:

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

remove parentheses:

$\displaystyle y + 6= - \frac{1}{2} (x - 6)$

distribute:

$\displaystyle y + 6= - \frac{1}{2} x + 3$

cancel 6 from both sides:

$\displaystyle y = - \frac{1}{2} x - 3$

hence, the equation of line k is y=-½x-3