Find the equation of the line whose y-intercept is - 3 and which is perpendicular to the line joining the points (- 2, 3) and (4, - 4) ?

CONCEPT:

Two point form:

The equation of a line passing through the points (x₁, y₁) and (x₂, y₂) is given as:

Slope - Point Form:

The equation of a line passing through the point (x1, y1) and having the slope ‘m’ is given as: y – y1 = m ⋅ (x – x1)

Note: If two lines are perpendicular then product of their slopes is – 1, i.e m1 ⋅ m2 = -1

CALCULATION:

Here, we have to find the the  equation of the line joining the points (- 2, 3) and (4, - 4) and having y-intercept - 3.

First let's find the equation of the line joining the points (- 2, 3) and (4, - 4).

⇒ 

So, the equation of line is: 7x + 6y - 32 = 0

The given equation 7x + 6y - 32 = 0 can be re-written as:  

By comparing the above with y = mx + c, we get slope of the give line i.e 

let slope of the normal be m2 

∵ The angle between the line and the required normal is 90°

As we know that, if two lines are perpendicular then product of their slopes is – 1, i.e m1 ⋅ m2 = -1

⇒ 

As we know that, the equation of a line passing through the point (x1, y1) and having the slope ‘m’ is given as: y – y1 = m ⋅ (x – x1)

∵ The normal has y-intercept - 3 i.e it passes through the point (0, - 3)

So, the equation of the normal whose slope is m2 = 6/7 and passes through the point (0, - 3)

⇒ 

So, the equation of the required normal is: 6x - 7y - 21 = 0

Hence, option C is the correct answer