To find the equation of the line provided the slope and a point:

y − y1 = m(x−x1) where(x1, y1)is the point given andmmis the slope.

That is quite straightforward and the required equation of the line is

y − 2 = 3/4 (x - 3),

rearranging we get4y − 3x + 1 = 0.

Now to the ‘point at a distance of 5’ part:

Imagine your required point on the line is(p,q), and(x1,y1)is a point given on the line, we have

q − y1/sinθ = p − x1/cosθ = r, where r is the distance between the two points on the line andθ, is the angle formed by the line with the x-axis (in the anti-clockwise direction).

The above said is theequation of the line in parametric form, where the parameter is the numerical distance between ANY two points on the line.

3/4 = m = tanθ = sinθ/cosθ, so we havesinθ=3/5

i.eq − 2/3/5 = p − 3/4/5 = 5

Orq − 2/3 = p − 3/4 = 5/5 = −1

Solving again we get,q = −1andp = −1

Curious to know, if both these points lie on our line, please go ahead and substitute in the equation of the line4y−3x+1=0.

(7, 5)and(−1, −1) both lie on our line and both are at a distance of55 units from(3. 2)

P.S. The parametric form of the equation of the line can be proven by using the idea from Ratios

a/b = c/d = K thena2 + b2/c2 + d2 = K2