Find the equation of the straight line passing through the point (3,5)
which is perpendicular to the line
y = 3x + 2

Step-by-step explanation:

y = 3x + 2

Slope m = 3

slope of the line perpendicular to this line = -3

Point (3,5)

y- y1 = m(x-x1)

y - 5 = -3(x-3)

y - 5 = -3*x-3 * -3

y - 5 = -3x + 9

     y = -3x + 9 +5

     y = -3x + 14

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.