Question - Equation of Perpendicular Line Passing Through a Point
Solution:
To find the equation of the line h, which is perpendicular to line g, and passes through the point (3, −9), we first need to determine the slope of line g.The equation for line g is given by:\[ y = \frac{3}{10}x - \frac{8}{10} \]The slope of line g is the coefficient of x, which is $$\frac{3}{10}$$.Two lines that are perpendicular to each other have slopes that are negative reciprocals of one another. So, the slope (m) of line h will be:\[ m = -\frac{1}{\text{slope of g}} = -\frac{1}{\frac{3}{10}} = -\frac{10}{3} \]Now we know the slope of line h, we can use the point-slope form of the equation of a line to find its equation. The point-slope form is:\[ y - y_1 = m(x - x_1) \]Plugging in our values for the slope and the coordinates of the given point (3, −9), we get:\[ y - (-9) = -\frac{10}{3}(x - 3) \]\[ y + 9 = -\frac{10}{3}x + 10 \]To write this in slope-intercept form, which is $$ y = mx + b $$, we solve for y:\[ y = -\frac{10}{3}x + 10 - 9 \]\[ y = -\frac{10}{3}x + 1 \]Here $$ b $$, the y-intercept, is 1. The numbers in the equation are already simplified proper fractions, improper fractions, or integers. So, the equation for line h in slope-intercept form is:\[ y = -\frac{10}{3}x + 1 \]
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