Given: Equation of line is 3x + 2y = 8
ere, we have to find the equation of the normal to the given line and passing through the midpoint of the line joining the points (6, 4) and (4, - 2).
First let's find the midpoint of he line joining the points (6, 4) and (4, - 2).
⇒(,)=(5,1) We can re-write the given equation of line as:
y=−.x+4 Now by comparing the above equation 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) So, the equation of the normal whose slope is
m2= and passing through the point
(5,1) ⇒y−1=.(x−5) So, the equation of the required normal is: 2x - 3y - 7 = 0
Hence, option C is the correct answer