Line Q is parallel to the line represented by the equation 5x-4y=10 line Q passes thrpugh the point (12,-7)

y = (5/4)x - 22

Step-by-step explanation:

Parallel lines always have the same slope. That means if you rework this equation into y=mx+b form it will be easier to see the slope.

-4y = -5x + 10

y = (5/4)x - (5/2)

Based on this we know the slope has to be (5/4)

Then, with the given point values we can solve for b

-7 = (5/4)(12) + b

-7 = 15 + b

b = -22

so the equation for line Q is y = (5/4)x -22

repeat the question

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⭐