Answer: (–4, –8)
Step-by-step explanation:
Let the coordinates of common point of the given lines are (x,y),
Thus, the slope of the line passes through the points (a,b) and R(4,2) is,
[tex]m_1 = \frac{2-b}{4-a}[/tex]
Again, the slope of the line passes through two points P(-6,4) and Q(4,-4),
[tex]m_2 = \frac{-4-4}{4-(-6)} = \frac{-8}{4+6} = \frac{-8}{10} = \frac{-4}{5}[/tex]
Since, both lines are perpendicular to each other,
⇒ [tex]m_1\times m_2=-1[/tex]
⇒ [tex]\frac{2-b}{4-a}\times \frac{-4}{5} = - 1[/tex]
⇒ [tex]\frac{4(2-b)}{5(4-a)} = 1[/tex]
⇒ [tex]\frac{8-4b}{20-5a} =1[/tex]
⇒ [tex]8-4b = 20 - 5a[/tex]
⇒ [tex]5a-4b=12[/tex] ---------(1)
For, a = -6 and b = 10,
[tex]5\times -6-4\times 10=-70\neq 12[/tex]
For, a = -4 and b = -8,
[tex]5\times -4-4\times -8=12= 12[/tex]
For, a = 0 and b = -1,
[tex]5\times 0-4\times-1=4\neq 12[/tex]
For, a =2 and b = 4,
[tex]5\times 2-4\times 4=-6\neq 12[/tex]
⇒ Second Option is correct.
