Given:
A line passes through (-5,-3) and perpendicular to [tex]y=-\dfrac{9}{4}x-5[/tex].
To find:
The equation of the line.
Solution:
We have,
[tex]y=-\dfrac{9}{4}x-5[/tex]
On comparing this equation with slope intercept form, i.e., [tex]y=mx+b[/tex], we get
[tex]m_2=-\dfrac{9}{4}[/tex]
It means, slope of this line is [tex]-\dfrac{9}{4}[/tex].
Product of slopes of two perpendicular lines is always -1.
[tex]m_1\times m_2=-1[/tex]
[tex]m_1 \times \left(-\dfrac{9}{4}\right)=-1[/tex]
[tex]m_1=\dfrac{4}{9}[/tex]
Slope of required line is [tex]\dfrac{4}{9}[/tex] and it passes through the point (-5,-3). So, the equation of the line is
[tex]y-y_1=m(x-x_1)[/tex]
where, m is slope.
[tex]y-(-3)=\dfrac{4}{9}(x-(-5))[/tex]
[tex]y+3=\dfrac{4}{9}(x+5)[/tex]
[tex]9(y+3)=4(x+5)[/tex]
[tex]9y+27=4x+20[/tex]
[tex]27-20=4x-9y[/tex]
[tex]7=4x-9y[/tex]
Therefore, the equation of required line is [tex]4x-9y=7[/tex].
