Aline passes through the point (6,-3) and has a slope of -3/2 write an equation in slope intercept form

The equation of the line with a slope of -3/2 passing through the point (6, -3) in slope-intercept form is expressed as

The formula for calculating the equation of a line in slope-intercept form is expressed as;

m is the slope

b is the y-intercept

Given the following parameters

m = -3/2

(x0, y0) =(6, -3)

Get the equation in the point-slope form first

Hence the equation in slope-intercept form is expressed as

Learn more here: link

y = - x + 6

Step-by-step explanation:

the equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y-intercept )

here m = - , hence

y = - x + c ← is the partial equation

to find c substitute (6, - 3 ) into the partial equation

- 3 = - 9 + c ⇒ c = - 3 + 9 = 6

y = - x + 6 ← equation in slope- intercept form

it's both, now can you answer my most recent question?

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐