Find the equation of the straight line passing throughthe point (3,5) which is perpendicular to the line y=3x+2 explain

Y=-1/3x+2 The negative 1/3 makes it perpendicular
5=(3)(-1/3)+b Plug in x and y
5=-1+b
6=b
Y=-1/3x+6

A rule in coordinate geometry you have to know 1..When two lines are perpendicular to each other then the product of their gradient is - 1...i.e(m¹×m²=-1)
2.. When two lines are parallel to each other their gradient is the same I.e m¹ =m²...

This question will make use of the first rule...

The second straight line was given to be y =3x+2.. Generally a straight line has the formula y=mx+c(where m=gradient, c=intercept)

So in y=3x+2(the gradient here is 3)

Now from the first rule the product of this line and the line of coordinates (3,5) is - 1..
:- m1*m2=-1
3×m2=-1
m2 = - 1/3this is the gradient of the line with coordinate (3,5)

Gradient = (y-y₁)/(x-x₁)

-1/3 = (y-5)/(x-3)

Cross multiply you'll have

-1(x-3)=3(y-5)
-x+3=3y-15

3y+x-15-3=0

3y+x-18=0 Checked

Or 3y+x=18

Hope this helped?