Can someone help me solve this?

(a) y = -3/5 x + 13/5

(b) y = 5/3 x + 1/3

Step-by-step explanation:

(a) The slope of the tangent line is dy/dx.  Use implicit differentiation:

x² + y² + 4x + 6y − 21 = 0

2x + 2y dy/dx + 4 + 6 dy/x = 0

2x + 4 + (2y + 6) dy/dx = 0

x + 2 + (y + 3) dy/dx = 0

(y + 3) dy/dx = -(x + 2)

dy/dx = -(x + 2) / (y + 3)

At the point (1, 2), the slope is:

dy/dx = -(1 + 2) / (2 + 3)

dy/dx = -3/5

Using point-slope form of a line:

y − 2 = -3/5 (x − 1)

Simplifying to slope-intercept form:

y − 2 = -3/5 x + 3/5

y = -3/5 x + 13/5

(b) The normal line is perpendicular to the tangent line, so its slope is 5/3.  It also passes through the point (1, 2), so point-slope form of the line is:

y − 2 = 5/3 (x − 1)

Simplifying to slope-intercept form:

y − 2 = 5/3 x − 5/3

y = 5/3 x + 1/3

1. 3

2. 4

3. 1

hope this ! have a great day.

-wfz

uhhh idk u yolo?