#### To determine

**To find:** The equation of the normal to the curve which is parallel to the given line.

#### Answer

The equation of normal line to the curve that is parallel to the line 2x+y=1 is y=−2x+3_.

#### Explanation

**Given:**

The equation of the curve is y=x and the line is 2x+y=1.

**Derivative rules:**

Power Rule: ddx(xn)=nxn−1

**Formula used:**

The equation of the normal line at (x1,y1) is, y−y1=m(x−x1) (1)

where, *m* is the slope of the normal line at (x1,y1) .

**Calculation:**

Obtain the slope of the normal line.

Given line rewritten as, y=−2x+1 and its slope is −2.

Since the required equation of the normal line is parallel to the given line, the slope of the normal line is equal to the slope of the given line.

Therefore, the slope of the normal line is −2.

Obtain the slope of the tangent line by using the slope of the normal line.

Here, the tangent line is perpendicular to the normal line.

That is, if m1 and m2 are the slopes of tangent line and normal line, then m1m2=−1.

Therefore, the slope of the tangent line is −1(−2)=12.

Obtain the normal line point on the curve.

The derivative of curve *y* is as follows,

dydx=ddx(x)=ddx(x12)

Apply the power rule and simplify the expression,

dydx=12x12−1=12x−12=12x12

Therefore, the slope of the tangent to the curve y=x is dydx=12x .

Since dydx=12 and dydx=12x, the value of *x* is computed as follows.

12=12xx=1x=12x=1

Thus, the value of x=1.

Substitute 1 for *x* in y=x,

y=1=1

Thus, the value of y=1.

Therefore, the normal line passes through the point (1, 1).

Obtain the equation of normal line.

Substitute (1,1) for (x1,y1) and -2 for *m* in equation (1),

y−1=−2(x−1)y−1=−2x+2y=−2x+3

Therefore, the equation of the normal line to the curve that is parallel to the line is y=−2x+3_.