Determine the equation for a line that passes through the point M(-7, 3) and perpendicular to 7x − 2y + 3 = 0. Show your complete work.

y = 2x/7 + 5

Step-by-step explanation:

the first line is defined by

7x - 2y + 3 = 0

in the most common way to write this we need an equation like y = ...

so, we isolate the y-expression in the equation :

2y = 7x + 3

y = (7x + 3) / 2 = 7x/2 + 3/2

the factor of x is the slope of the line, which is expressed by the y/x ratio indicating how many units y will change, when x changes e.g. 1 unit.

7/2 means that y will change by 7 units for every change in x by 2 units.

the slope of a perpendicular line is inverse to the slope of the original line, which means x and y are simply trading places (original x / original y).

so, the perpendicular slope of 7/2 is 2/7 (when x changes by 7 units, then y changes by 2 units).

that means the equation for the perpendicular line is something like

y = 2x/7 + c

the c we determine by using the coordinates of the given point (x and y values).

3 = 2×(-7)/7 + c = 2×(-1) + c = -2 + c

c = 5

so, the complete equation of the perpendicular line is

y = 2x/7 + 5

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.

problem:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

f(5/2) = 10/3

i hope this !