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Question: Find distance of point A(2, 3) measured parallel to the line $x - y = 5$ from the line $2x + y + 6 =...

Find distance of point A(2, 3) measured parallel to the line x−y=5x - y = 5 from the line 2x+y+6=02x + y + 6 = 0.

Answer

1323\frac{13\sqrt{2}}{3}

Explanation

Solution

Core Solution:

  1. Direction of the parallel line: The line x−y=5x - y = 5 has a direction vector (1,1)(1, 1).
  2. Parametric equation of the line through A: The line passing through A(2, 3) and parallel to x−y=5x - y = 5 can be represented parametrically as: x=2+λx = 2 + \lambda y=3+λy = 3 + \lambda
  3. Point of intersection: To find where this line intersects 2x+y+6=02x + y + 6 = 0, substitute the parametric equations into the line equation: 2(2+λ)+(3+λ)+6=02(2 + \lambda) + (3 + \lambda) + 6 = 0 4+2λ+3+λ+6=04 + 2\lambda + 3 + \lambda + 6 = 0 3λ+13=03\lambda + 13 = 0 λ=−133\lambda = -\frac{13}{3}
  4. Coordinates of intersection point (B): Substitute λ=−133\lambda = -\frac{13}{3} back into the parametric equations: xB=2+(−133)=6−133=−73x_B = 2 + (-\frac{13}{3}) = \frac{6 - 13}{3} = -\frac{7}{3} yB=3+(−133)=9−133=−43y_B = 3 + (-\frac{13}{3}) = \frac{9 - 13}{3} = -\frac{4}{3} So, the intersection point B is (−73,−43)(-\frac{7}{3}, -\frac{4}{3}).
  5. Distance between A and B: Use the distance formula for points A(2, 3) and B(−73,−43)(-\frac{7}{3}, -\frac{4}{3}): d=((x2−x1)2+(y2−y1)2)d = \sqrt{((x_2 - x_1)^2 + (y_2 - y_1)^2)} d=((−73−2)2+(−43−3)2)d = \sqrt{((-\frac{7}{3} - 2)^2 + (-\frac{4}{3} - 3)^2)} d=((−73−63)2+(−43−93)2)d = \sqrt{((-\frac{7}{3} - \frac{6}{3})^2 + (-\frac{4}{3} - \frac{9}{3})^2)} d=((−133)2+(−133)2)d = \sqrt{((-\frac{13}{3})^2 + (-\frac{13}{3})^2)} d=(1699+1699)d = \sqrt{(\frac{169}{9} + \frac{169}{9})} d=(2×1699)d = \sqrt{(\frac{2 \times 169}{9})} d=169×29d = \frac{\sqrt{169} \times \sqrt{2}}{\sqrt{9}} d=1323d = \frac{13\sqrt{2}}{3}

Answer: The distance is 1323\frac{13\sqrt{2}}{3} units.