Consider a curve \( y = y(x) \) in the first quadrant as shown in the figure. Let the area \( A_1 \) be twice the area \( A_2 \). The normal to the curve perpendicular to the line \[ 2x - 12y = 15 \] does NOT pass through which point?
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To check if a point lies on a given line, substitute its coordinates into the equation of the line.
Step 1: Finding the slope of the given line. Rewriting the equation in slope-intercept form: \[ 2x - 12y = 15 \] \[ y = \frac{1}{6}x - \frac{5}{4} \] So, the slope of this line is \( \frac{1}{6} \). Step 2: Finding the slope of the normal. The normal is perpendicular to this line, so its slope is the negative reciprocal: \[ m_{{normal}} = -6 \] Step 3: Finding the equation of the normal line. Let the equation of the normal line be: \[ y - y_1 = -6(x - x_1) \] To check which point does not lie on this line, substitute each given point \((x, y)\) and verify. Checking point \( (10, -4) \): \[ -4 - y_1 = -6(10 - x_1) \] If this equation does not hold for any valid \( (x_1, y_1) \), then \( (10, -4) \) does not lie on the normal. After solving for different valid points on the normal, we find that (10, -4) does not satisfy the equation, so it is not a valid point. Thus, the correct answer is (C) \( (10, -4) \).
