Question

1. find an equation of the line perpendicular to the line with equation ( y - 2 = -5(x - 1) ) passing through the point ( (-10, 7) ).

Explanation:

Step1: Find the slope of the given line

The given line is in point - slope form \(y - y_1=m(x - x_1)\), where \(m\) is the slope. For the line \(y - 2=-5(x - 1)\), the slope \(m_1=- 5\).
If two lines are perpendicular, the product of their slopes \(m_1\times m_2=-1\). Let the slope of the perpendicular line be \(m_2\). Then \(-5\times m_2=-1\), so \(m_2=\frac{1}{5}\).

Step2: Use point - slope form to find the equation of the new line

The point - slope form of a line is \(y - y_0=m(x - x_0)\), where \((x_0,y_0)=(-10,7)\) and \(m = \frac{1}{5}\).
Substitute these values into the formula: \(y - 7=\frac{1}{5}(x+ 10)\).
Expand the right - hand side: \(y - 7=\frac{1}{5}x + 2\).
Add 7 to both sides of the equation: \(y=\frac{1}{5}x+2 + 7\), so \(y=\frac{1}{5}x + 9\). We can also write it in standard form: \(x-5y+45 = 0\) or in point - slope form as \(y - 7=\frac{1}{5}(x + 10)\).

Answer:

The equation of the perpendicular line is \(y=\frac{1}{5}x + 9\) (or \(y - 7=\frac{1}{5}(x + 10)\) or \(x-5y+45 = 0\))