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TS EAMCET · Maths · Application of Derivatives

If the point \(\mathrm{P}\left(x_1, y_1\right)\) 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 \(3 x+4 y-2=0\) is

  1. A \(\frac{16}{5}\)
  2. B 4
  3. C 1
  4. D \(\frac{7}{5}\)
Verified Solution

Answer & Solution

Correct Answer

(C) 1

Step-by-step Solution

Detailed explanation

Tangent slope \(m_{tan} = 2x-1\). Line slope \(m_{line} = 1\). \(2x-1 = 1 \Rightarrow 2x=2 \Rightarrow x_1=1\). \(y_1 = (1)^2-1+1 = 1\). \(\mathrm{P}(1,1)\). Distance \(D = \frac{|3(1)+4(1)-2|}{\sqrt{3^2+4^2}}\). \(D = \frac{|3+4-2|}{\sqrt{9+16}}\).…