Question

If the straight lines 2x+3y-1=0, x+2y-1=0 and ax+by=1 form a triangle with origin as orthocentre then (a,b) is given by

(a)(6,4) (b)(-3,3) (c)(-8,8) (d)(0,7)

Answer 1

Given:

• (Refer to the attached diagram)

• Equation of the sides of triangle are:
• 2 x + 3 y - 1 = 0
• x + 2 y - 1 = 0
• a x + b y - 1 = 0
• Orthocentre is origin i.e. (0,0)

To find:

• (a,b)

Solution:

• Line AB and line AC pass through A and therefore form a pair of lines at point A.
• Hence, line AO = 2x+3y-1 + ω( x + 2y -1) = 0
• Where ω is a constant.
• Since, AO passes through origin, we substitute x=0 and y = 0 in the equation of AO.
• We get ω = -1.
• Substitute in original equation of AO, we get equation of Ao as
• x + y = 0
• Now, line AO and line BC are perpendicular to each other.
• Hence Product of slopes of the two perpendicular lines is -1.
• Slope of AO = -1 and Slope of BC = -a/b
• ∴ -1 x -(a/b) = -1
• a = - b
• Similarly, equation of line BO will be
• (2x + 3y -1) + β ( ax + by - 1) = 0       where β is a constant
• Substituting a = -b equation becomes,
• x(2 + aβ) + y(3 - aβ) + (-1 - β) = 0
• This line BO passes through origin. Substituting x = 0 and y = 0 we get,
• β = -1.
• Substituting in equation of BO we get equation of BO as
• x (2-a) + y (3 + a) = 0
• Slope of this line = -(2-a)/3+a
• Line BO and line AC.
• Hence, product of their slopes is -1
• Slope of BO is already found and slope of AC = -1/2
• $\frac{-(2-a)}{3+a} * \frac{-1}{2} = -1$
• ∴ 2 - a = 6 - 2 a
• a = -8
• But a = -b
• ∴ b = 8.

Answer:

(a,b) is given by (-8,8).