Find a equation of a line that is perpendicular to y=-4x+3 and passes through the point (4,-1)

y = (-1/4)x

Step-by-step explanation:

the factor of x is in a line equation the slope of the line.

it residents actually the ratio of y/x.

in our original equation that means that for every change of x by 1 unit y changes by 4 units (actually -4 units) => -4/1 ratio.

a perpendicular line just reverses that ratio.

now, x had to change by 4 units for y to change by 1 unit (actually -1 unit)

=> -1/4

so, our wanted line equation must be in the form of

y = (-1/4)x + c

c is the offset of the line from the (0,0) point in y direction.

so, what is c ?

now we use the information of the point that is stored to be in that line : (4,-1)

=> -1 = (-1/4)×4 + c

=> -1 = -1 + c

=> c = 0

therefore, the desired line is

y = (-1/4)x

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.

college