The equation of line lm is y = 5x + 4. write an equation of a line perpendicular to line lm in slope-intercept form that contains point (−3, 2). y = 5x + 13 y = negative 1 over 5x + 7 over 5 y = − 1 over 5x − 7 over 5 y = 5x − 17

Equation of LM is ,
y=5x+4
Comparing it with y=mx+c
slope of LM(m1)=5
Let m2 be slope of line perpendicular to LM ,then,
m1×m2=-1
m2=-1/m1
m2=-1/5
Now , the equation of straight line with slope m2 & passing through (-3,2) can be given in slope intercept form as,
↪m2=(y-y1)(x-x1)
↪-1/5=(y-2)/(x+3)
↪-x-3=5y-10
↪0=x+3+5y-10
↪x+5y-7=0
↪5y=-x+7
↪y=(-1/5)x+(7/5)➡2nd option