To find the value of \( x \) such that the line passing through points (1, 2) and (x, 5) is perpendicular to a line with a slope of -10, we need to use the properties of slopes of perpendicular lines.
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Identify the slope of the line with points (1, 2) and (x, 5): The slope \( m_1 \) of the line connecting points (1, 2) and (x, 5) is given by the formula: \[ m_1 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - 2}{x - 1} = \frac{3}{x - 1} \]
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Find the slope of the line that is required to be perpendicular: Given that the slope of the other line is -10, we know that if two lines are perpendicular, the product of their slopes is -1: \[ m_1 \cdot m_2 = -1 \] Here, \( m_2 = -10 \). So we have: \[ \frac{3}{x - 1} \cdot (-10) = -1 \]
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Solve for \( x \): This leads to the equation: \[ -\frac{30}{x - 1} = -1 \] Removing the negative signs: \[ \frac{30}{x - 1} = 1 \] Cross-multiplying gives: \[ 30 = x - 1 \] Solving for \( x \): \[ x = 30 + 1 = 31 \]
However, upon review, it appears my initial conditions should yield a scenario where we equate \( m_1 \) to \(\frac{1}{10}\) as the slope perpendicular to the given slope of -10 (as slopes of perpendicular lines are negative reciprocals). Thus, the initial equation can be restated correctly as:
\[ \frac{3}{x - 1} = \frac{1}{10} \]
Redoing the calculation here gives us: Cross-multiplying results in: \[ 3 \cdot 10 = (x - 1) \cdot 1 \] \[ 30 = x - 1 \] Thus, \[ x = 31 \]
From this, we arrive at your understanding that the value of \( x \) does not match those provided as options.
Based on further calculations or consideration, if we want it to yield a coefficient from one of the given choices, you may review options for a specific value perhaps as related to error detection vs slope, however for clarity, the correct \( x \) for the derived line towards -10 slope is 31.
In essence: If the choice is to retrieve explicitly one of the initial provided options, they may need adjusting reliably based on slope recalibrations.