Question

Through: (-5, -3), perp. to y=-9/4x-5

Answer 1

Given:

A line passes through (-5,-3) and perpendicular to $y=-\dfrac{9}{4}x-5$.

To find:

The equation of the line.

Solution:

We have,

$y=-\dfrac{9}{4}x-5$

On comparing this equation with slope intercept form, i.e., $y=mx+b$, we get

$m_2=-\dfrac{9}{4}$

It means, slope of this line is $-\dfrac{9}{4}$.

Product of slopes of two perpendicular lines is always -1.

$m_1\times m_2=-1$

$m_1 \times \left(-\dfrac{9}{4}\right)=-1$

$m_1=\dfrac{4}{9}$

Slope of required line is $\dfrac{4}{9}$ and it passes through the point (-5,-3). So, the equation of the line is

$y-y_1=m(x-x_1)$

where, m is slope.

$y-(-3)=\dfrac{4}{9}(x-(-5))$

$y+3=\dfrac{4}{9}(x+5)$

$9(y+3)=4(x+5)$

$9y+27=4x+20$

$27-20=4x-9y$

$7=4x-9y$

Therefore, the equation of required line is $4x-9y=7$.