Question

Find the equation of line passing through point (4,5) and perpendicular to the line 7x-5y=420

Answer 1

7x - 5y = 420

y = 7x/5 - 420/5

y = 7x/5 - 84

m1=7/5

Therefore m2 = -5/7

y-y1=m(x-x1)

y-4=-5/7(x-5)

7y-28=5x-25

5x-7y-3=0.

Answer 2

Answer:

The required equation is 5x + 7y = 55

Step-by-step explanation:

Given data  

given line  7x - 5y = 420 _(1)  

here we need to find equation of the line passing through point (4, 5) and perpendicular to

⇒ given line (1) can be written as  5y = 7x - 420

                                                         y = $\frac{7}{5}$ x - 84

⇒ which is in the form of y = mx + c [ here m is slope of the line]

⇒ compare y = $\frac{7}{5}$x +c  with y = mx + c

⇒ Slope of line (1)  m₁ = 7/5  

Let m₂ be the slope of the required line

Note : when two lines are perpendicular to each other then product of their slope is -1  

⇒  m₁ m₂  = -1     ⇒ $\frac{7}{5}$ m₂ = -1     ⇒ m₂ = - 5/7  

⇒ we know that the equation of the line with slope m₁, passing through a point  (x₁, y₁)   = y - y₁ = m₁ ( x - x₁)  

⇒ Equation of the line passing through (4, 5) with slope $-\frac{5}{7}$  

   ⇒  (y - 5)  = $-\frac{5}{7}$ (x - 4)  

   ⇒  7(y -5) = -5(x - 4)

   ⇒  7y - 35 = -5x +20

   ⇒ 5x + 7y = 55