Question

A line Li passes through point (1, 2) and has a gradient of 5. Another line L2, is
perpendicular to L, and meets it at a point where x=4. Find the equation for Le in the
form y = mx +c.

Answer 1

Answer:

The equation of line Le is 5 y + x + 89 = 0

Step-by-step explanation:

Given as :

The line Li passes through point (1,2)

The slope of line Li (m) = 5

So, equation of line y = m x + c

or,  2 = 5 × 1 + c

Or, c = 2 - 5 = -3

∴ Eq of line Li is

y = 5 x - 3

Another line Le is perpendicular to line Li

Let the slope of line Le = M

The line Le meet at points where x = 4

So, for perpendicular property

Products of slopes = - 1

So, m × M = - 1

or, M = - $\frac{1}{m}$

Or. M = - $\frac{1}{5}$

Now equation of line Le is y = M x + c

∵ Line Le meet the line Li at x = 4

So, y = 5 × 4 - 3

I.e  y = 20 - 3

Or,  y = 17

Now, equation of line Le with slope - $\frac{1}{5}$ and passes through points ( 4 , 17 ) is

y = M x + c

or, 17 =  - $\frac{1}{5}$ × 4 + c

or, 17 +   $\frac{4}{5}$ = c

So, c =   $\frac{85+4}{5}$

Or, c =  - $\frac{89}{5}$

∴ Equation of line Le is y =  - $\frac{1}{5}$ x - $\frac{89}{5}$

ie.  5 y = - x - 89

or, 5 y + x + 89 = 0  

Hence The equation of line Le is 5 y + x + 89 = 0   answer