Question

question
line t passes through the points $(-6,-2)$ and $(1,-2)$. what is the equation for
the line which is perpendicular to t and passes through the point $(-2,5)$?
answer attempt 1 out of 2
$x = 1$
$x=-2$
$y = 1$
$y=-2$
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Explanation:

Step1: Find the slope of line \(t\)

Use the slope formula \(m=\frac{y_2 - y_1}{x_2 - x_1}\). Given points \((-6,-2)\) and \((1,-2)\), then \(m=\frac{-2-(-2)}{1 - (-6)}=\frac{0}{7}=0\).

Step2: Determine the slope of the perpendicular line

If two lines are perpendicular, the product of their slopes is \(- 1\). Let \(m_1\) be the slope of line \(t\) (\(m_1 = 0\)) and \(m_2\) be the slope of the perpendicular line. Since \(m_1\times m_2=-1\), and \(m_1 = 0\), the perpendicular line is a vertical line (undefined slope in the traditional \(y = mx + b\) sense, but vertical lines have the form \(x = k\)).

Step3: Find the equation of the perpendicular line

The perpendicular line passes through the point \((-2,5)\). For a vertical line \(x = k\), substituting \(x=-2\) (from the point \((-2,5)\)) into \(x = k\), we get \(x=-2\).

Answer:

\(x=-2\)