Question

If the point P($x_1, y_1$) lying on the curve $y = x^2-x+1$ is the closest point to the line $y = x-3$ then the perpendicular distance from P to the line $3x+4y-2=0$ is

• A. $16/5$
• B. 4
• C. 1
• D. $7/5$

Hint:

To find the shortest distance between a curve and a line, find the point on the curve where the tangent is parallel to the line. The shortest distance is then the perpendicular distance from this point to the line.

Answer 1

The Correct Option is C

The point on the curve $y=f(x)$ that is closest to a line $y=mx+c$ is the point where the tangent to the curve is parallel to the line.
The given curve is $y = x^2-x+1$.
The given line is $y=x-3$, which has a slope of $m_L=1$.
The slope of the tangent to the curve is given by its derivative, $\frac{dy}{dx}$.
$\frac{dy}{dx} = \frac{d}{dx}(x^2-x+1) = 2x-1$.
For the tangent to be parallel to the line, their slopes must be equal.
$\frac{dy}{dx} = m_L \implies 2x-1 = 1$.
$2x=2 \implies x=1$.
This is the x-coordinate of the point P. To find the y-coordinate, substitute $x=1$ into the curve's equation:
$y = (1)^2 - (1) + 1 = 1$.
So, the point P is $(1,1)$.
Now, we need to find the perpendicular distance from the point P(1,1) to the line $3x+4y-2=0$.
The formula for the perpendicular distance from a point $(x_1, y_1)$ to a line $Ax+By+C=0$ is $d = \frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}$.
$d = \frac{|3(1)+4(1)-2|}{\sqrt{3^2+4^2}} = \frac{|3+4-2|}{\sqrt{9+16}} = \frac{|5|}{\sqrt{25}} = \frac{5}{5} = 1$.