Line p passes through points (2,13) and (9,8). line q is perpendicular to p what is the slope of line q?

to find the slope of P, use the formula (y2-y1)/(x2-x1)

(8-13)/(9-2) gives you -5/7

a perpendicular line has the opposite reciprocal slope so it is positive 7/5 or 1 2/5 

< 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⭐

20%

step-by-step explanation: