Here, we have to find the the equation of the line perpendicular to the line 3x + 5y = 4 and having x-intercept - 3.
The given equation
3x+5y=4 can be re-written as:
y=−.x+ By comparing the above with y = mx + c, we get slope of the give line i.e
m1=− 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 ⇒−.m2=−1 ⇒m2= 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 x-intercept - 3 i.e it passes through the point (- 3, 0)
So, the equation of the normal whose slope is
m2=5∕3 and passes through the point (3, 0)
⇒y−0=.(x+3) So, the equation of the required normal is:
5x−3y+15=0 Hence, option D is the correct answer.