Question

Find the equation of the line passing through a point A(2, -3) and perpendicular to the line having the parametric equations: 5 = 3 + 4 and 3 = 4 − 6​

Answer 1

Answer:

The resultant equation is : 4x - 3y + 1 =0.

Step-by-step explanation:

Given point is : (2,3)

The given equation is : 3x + 4y - 5 = 0.

So, first we find the slope of the equation,

3x + 4y - 5 = 0

4y = 5 - 3x

So, y = 5/4 - 3x/4

The Slope is : -3/4

By the condition of perpendicularity,

M1 × M2 = -1

-3/4 × M2 = -1

M2 = 4/3

The Slope of perpendicular condition ( M2) is : 4/3.

Given point ( X1 , Y1 ) = (2,3).

On finding the equation ,

So, we know that : Y - Y1 = M2 ( X - X1 )

y - 3 = 4/3 ( x - 2 )

3y - 9 = 4x - 8

-9 +8 = 4x - 3y

4x - 3y = -1

4x - 3y + 1 = 0

The resultant equation is : 4x - 3y + 1 =0.