Given:
The equation of a line is:
[tex]y=-\dfrac{5}{7}x+2[/tex]
A line passes through the point (-5,-3) and perpendicular to the given line.
To find:
The equation of the line.
Solution:
Slope intercept form of a line is:
[tex]y=mx+b[/tex] ...(i)
Where, m is the slope and b is the y-intercept.
We have,
[tex]y=-\dfrac{5}{7}x+2[/tex] ...(ii)
On comparing (i) and (ii), we get
[tex]m=-\dfrac{5}{7}[/tex]
We know that the product of slopes of two perpendicular lines is always -1.
[tex]m_1\times m_2=-1[/tex]
[tex]-\dfrac{5}{7}\times m_2=-1[/tex]
[tex]m_2=\dfrac{7}{5}[/tex]
Slope of the required line is [tex]\dfrac{7}{5}[/tex] and it passes through the point (-5,-3). So, the equation of the line is:
[tex]y-y_1=m_2(x-x_1)[/tex]
[tex]y-(-3)=\dfrac{7}{5}(x-(-5))[/tex]
[tex]y+3=\dfrac{7}{5}(x+5)[/tex]
Using distributive property, we get
[tex]y+3=\dfrac{7}{5}(x)+\dfrac{7}{5}(5)[/tex]
[tex]y+3=\dfrac{7}{5}x+7[/tex]
[tex]y=\dfrac{7}{5}x+7-3[/tex]
[tex]y=\dfrac{7}{5}x+4[/tex]
Therefore, the equation of the line is [tex]y=\dfrac{7}{5}x+4[/tex]. Hence, option A is correct.
