Find the equation of the line that contains the given point and is parallel to the given line. Write the equation in slope-intercept form, if possible. (-6,8); 4x-3y=5

y = 4x/3 + 16

Step-by-step explanation:

Since the point new line is parallel to the line 4x - 3y = 5, so it must have the same slope. We now write this equation in slope-intercept form

4x - 3y = 5

-3y = -4x + 5

y = 4x/3 - 5/3

Its slope is the coefficient of x. So, its slope is 4/3 which will be the slope of the new parallel line which passes through the point (-6, 8).

To find the equation of this line, we use

(y - y₀)/(x - x₀) = m where x₀, y₀ are the points which the line passes through and m = slope of line. Given that ( x₀, y₀) = (-6, 8) and m = 4/3, we substitute these values into the equation. We have

(y - y₀)/(x - x₀) = m

(y - 8)/(x - (-6)) = 4/3

(y - 8)/(x + 6) = 4/3

cross-multiplying, we have

3(y - 8) = 4(x + 6)

Expanding the brackets, we have

3y - 24 = 4x + 24

collecting like terms, we have

3y = 4x + 24 + 24

3y = 4x + 48

In slope-intercept form, it is

y = 4x/3 + 48/3

y = 4x/3 + 16

< 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⭐