Rewrite the equation as y-2=-2/3(x-(-1)) to identify a point on the graph

  point (-1, 2)

Step-by-step explanation:

Your equation is shown here in point-slope form, so a point on the graph can be read from the given equation. (No further rewrite is necesary.)

The point-slope form is ...

  y -k = m(x -h)

Comparing this to your given equation ...

  y -2 = (-2/3)(x -(-1))

we see that ...

  k = 2, m = -2/3, h = -1

The point on the graph is (h, k) = (-1, 2).

26

step-by-step explanation:

< 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⭐