Respuesta :
Given:
Equation of line [tex]7x-4y=-12[/tex].
To find:
The equation of line that goes through the point ( − 21 , 2 ) and is perpendicular to the given line.
Solution:
The given equation of line can be written as
[tex]7x-4y+12=0[/tex]
Slope of line is
[tex]\text{Slope}=-\dfrac{\text{Coefficient of x}}{\text{Coefficient of y}}[/tex]
[tex]m_1=-\dfrac{7}{(-4)}[/tex]
[tex]m_1=\dfrac{7}{4}[/tex]
Product of slopes of two perpendicular lines is -1. So, slope of perpendicular line is
[tex]m_1m_2=-1[/tex]
[tex]m_2=-\dfrac{1}{m_1}[/tex]
[tex]m_2=-\dfrac{4}{7}[/tex] [tex][\because m_1=\dfrac{7}{4}][/tex]
Now, the slope of perpendicular line is [tex]m_2=\dfrac{4}{7}[/tex] and it goes through (-21,2). So, the equation of line is
[tex]y-y_1=m_2(x-x_1)[/tex]
[tex]y-2=-\dfrac{4}{7}(x-(-21))[/tex]
[tex]y-2=-\dfrac{4}{7}x-\dfrac{4}{7}(21)[/tex]
[tex]y-2=-\dfrac{4}{7}x-12[/tex]
[tex]y=-\dfrac{4}{7}x-12+2[/tex]
[tex]y=-\dfrac{4}{7}x-10[/tex]
Therefore, the required equation in slope intercept form is [tex]y=-\dfrac{4}{7}x-10[/tex].
