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6 a straight connecting street segment is built perpendicular to an exi…

Question

6 a straight connecting street segment is built perpendicular to an existing street with equation $y = \frac{2}{7}x + 3$. determine the equation of the line of the new street segment, which passes through point b(-1, -0.2).

Explanation:

Step1: Find perpendicular slope

The slope of the given line is $m_1=\frac{2}{7}$. For perpendicular lines, $m_2=-\frac{1}{m_1}=-\frac{7}{2}$.

Step2: Use point-slope form

Point-slope formula: $y - y_1 = m(x - x_1)$. Substitute $m=-\frac{7}{2}$, $x_1=-1$, $y_1=-0.2$:
$y - (-0.2) = -\frac{7}{2}(x - (-1))$

Step3: Simplify to slope-intercept form

First rewrite $-0.2$ as $-\frac{1}{5}$:
$y + \frac{1}{5} = -\frac{7}{2}(x + 1)$
Expand the right side:
$y + \frac{1}{5} = -\frac{7}{2}x - \frac{7}{2}$
Subtract $\frac{1}{5}$ from both sides:
$y = -\frac{7}{2}x - \frac{7}{2} - \frac{1}{5}$
Calculate the constant term: $-\frac{7}{2} - \frac{1}{5} = -\frac{35}{10} - \frac{2}{10} = -\frac{37}{10} = -3.7$

Answer:

$y = -\frac{7}{2}x - 3.7$ or $y = -\frac{7}{2}x - \frac{37}{10}$