Find the equation of the line with slope m=15 that contains the point (−10,−5).

Step-by-step explanation:

we know the slope is 15, and contains these points (-10,-5)

y=15x+b

Plug in x and y with those points.

-5=15(-10)+b

-5=-150+b

b=145

The equation is y=15x+145

< 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⭐

77

step-by-step explanation: