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Halle la ecuación de la recta que es perpendicular a la recta 
y=2x+1 y pasa por el punto 
(-2,7)

Halle la ecuación de la recta que es perpendicular a la recta y=2x+1 y=2 x+1 y pasa por el punto (2,7) (-2,7)

Full solution

Q. Halle la ecuación de la recta que es perpendicular a la recta y=2x+1 y=2 x+1 y pasa por el punto (2,7) (-2,7)
  1. Identify slope: Identify the slope of the given line y=2x+1y = 2x + 1. The slope (m)(m) is 22.
  2. Find perpendicular slope: Find the slope of the perpendicular line. The slope of the perpendicular line is the negative reciprocal of 22, which is 12-\frac{1}{2}.
  3. Use point-slope form: Use the point-slope form of the equation of a line, yy1=m(xx1)y - y_1 = m(x - x_1), where (x1,y1)(x_1, y_1) is the point (2,7)(-2, 7) and mm is 12-\frac{1}{2}.
  4. Substitute point and slope: Substitute the point (2,7)(-2, 7) and the slope 12-\frac{1}{2} into the point-slope form: y7=12(x+2)y - 7 = -\frac{1}{2}(x + 2).
  5. Simplify equation: Simplify the equation: y7=12x1y - 7 = -\frac{1}{2}x - 1.
  6. Convert to slope-intercept form: Add 77 to both sides to get the equation in slope-intercept form: y=12x+6y = -\frac{1}{2}x + 6.

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