Question

Find an equation of the line that is perpendicular to the line y=5x+6 and passes through the point (-2,5)

Answer 1

Answer:

$5y=-x+23$

Step-by-step explanation:

Since the equation of the line is perpendicular to the line y = 5x +6

we can use the formula:

$m_1m_2=-1$

where m1 is the slope of the line we have and,

m2 is the slope of the line we need

so here m1 is 5 as we compare it with the slope-intercept form our slope is:

$y=5x+6\\y=mx+b$

our m1 is 5 but we need m2 so plugging in the value we get m2 to be:

$m_1m_2=-1\\5m_2=-1\\m_2=-1/5$

and now we have our m2, now furthermore since the line passes through the point (-2 , 5) it means the point must lie on the line and satisfy it. We can use this point and the slope m2 and find out the equation by the point-slope form equation:

$y-y_1=m(x-x_1)\\y-5=-\frac{1}{5}(x-(-2))\\\\5(y-5)=(-1)(x+2)\\5y-25=-x-2\\5y=-x-2+25\\5y=-x+23$