Question

What is an equation bof the line that is perpendicular to y-4=2(x-6);and passes through the point (-3,-5)

Answer 1

Answer:

y+5=-1/2(x+3)

Step-by-step explanation:

as perpendicular,

compare that given eqn with y-y1=m(x-x1),

m1=2

then,

perpendicular case,

m1×m2=-1

m2=-1/2

now,

as the eqn passes through point (-3,-5),

we know,

y-y1=m(x-x1)

then putting value,

y+5=-1/2(x+3)

Answer 2

Answer:

y + 5 = - $\frac{1}{2}$(x + 3)

Step-by-step explanation:

The equation of a line in point- slope form is

y - b = m(x - a)

where m is the slope and (a, b) a point on the line.

y - 4 = 2(x - 6) is in this form with slope m = 2

Given a line with slope m then the slope of a line perpendicular to it is

$m_{perpendicular}$ = - $\frac{1}{m}$ = - $\frac{1}{2}$

Hence the equation passing through (- 3, - 5) is

y - (- 5) = - $\frac{1}2}$ (x - (- 3)), that is

y + 5 = - $\frac{1}{2}$(x + 3) ← equation of perpendicular line