If (a, b) is the centroid of the triangle formed by the lines 4x2 – 17xy + 4y2 = 0 and x + y – 5 = 0 and c is the numerical value of the area of the triangle, then a + b + c =
Answers
Solution :-
Pair of straight lines = 4x² - 17xy + 4y² = 0
→ 4x² - 17xy + 4y² = 0
→ 4x² - 16xy - xy + 4y² = 0
→ 4x(x - 4y) - y(x - 4y) = 0
→ (x - 4y)(4x - y) = 0
So, 2 lines are ,
→ x - 4y = 0 ----------(1)
→ 4x - y = 0 ----------(2)
and, third line is given ,
→ x + y - 5 = 0
→ x + y = 5 . ----------(3)
solving (1) and (2) , by adding both,
→ x - 4y + 4x - y = 0
→ 5x - 5y = 0
→ 5x = 5y
→ x = y
So,
→ (x1, y1) = (0,0)
Solving (2) and (3), by adding,
→ 4x - y + x + y = 0 + 5
→ 5x = 5
→ x = 1
putting value of x ,
→ 1 + y = 5
→ y = 5 - 1 = 4
So,
→ (x2,y2) = (1,4)
Solving (3) and (1) now, by subtracting (3) from (1) ,
→ (x - 4y) - (x + y) = 0 - 5
→ x - x - 4y - y = (-5)
→ (-5y) = (-5)
→ y = 1
putting value of y ,
→ x + 1 = 5
→ x = 4
So,
→ (x3,y3) = (4,1)
Therefore,
→ a = (x1 + x2 + x3)/3 = (0 + 1 + 4)/3 = (5/3)
→ b = (y1 + y2 + y3)/3 = (0 + 4 + 1)/3 = (5/3)
Now, we have to find area of the triangle.
Area of Triangle , when one of the coordinates is (0,0) can be solved directly by :-
- Let vertices of ∆ are are (0,0), (a, b), and (c, d) , than, area of ∆ will be = |(ad - bc)|/2
So,
→ Area of ∆ with coordinates (0,0), (1,4), and (4,1) is = |(1*1 - 4*4)|/2 = |(1 - 16)| / 2 = (15/2) unit². = c .
Hence,
→ a + b + c
→ (5/3) + (5/3) + (15/2)
→ (10/3) + (15/2)
→ (20 + 45)/6
→ (65/6) (Ans.)
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