If the line l 1 : 3 y - 2 x = 3 is the angular bisector of the lines l 2 : x - y + 1 = 0 and l 3 : x + y + 17 = 0 , then 2 + 2 - is equal to

Question

If the line $l_{1} : 3 y - 2 x = 3$ is the angular bisector of the lines $l_{2} : x - y + 1 = 0$ and $l_{3} : α x + β y + 17 = 0$ , then $α^{2} + β^{2} - α - β$ is equal to ______.

Smart Achievers · Accepted Answer

(348)

$Point of intersection of l_{1} : 3 y - 2 x = 3 and l_{2} : x - y + 1 = 0 is P (0, 1),$ $which lies on l_{3} : α x + β y + 17 = 0$

$β = - 17$

$Consider a random point Q (- 1, 0) on l_{2} : x - y + 1 = 0 .$ $Image of point Q about l_{1} : 2 x - 3 y + 3 = 0 is Q' (\frac{- 17}{13}, \frac{6}{13}) .$

$which can be calculated by formula,$ $\frac{x - (- 1)}{2} = \frac{y - 0}{- 3} = \frac{- 2 (- 2 + 3)}{13}$

$Now, Q' lies on l_{3} : α x + β y + 17 = 0$ $∴$ $α = 7$

$Now, α^{2} + β^{2} - α - β = 348$

Solution

Q.
If the line $l_{1} : 3 y - 2 x = 3$ is the angular bisector of the lines $l_{2} : x - y + 1 = 0$ and $l_{3} : α x + β y + 17 = 0$ , then $α^{2} + β^{2} - α - β$ is equal to ______.

Want Us to Email you About Special Offers & Updates?

Solution

Q. If the line l1:3y-2x=3 is the angular bisector of the lines l2:x-y+1=0 and l3:αx+βy+17=0, then α2+β2-α-β is equal to ______.

Want Us to Email you About Special Offers & Updates?

Q.
If the line $l_{1} : 3 y - 2 x = 3$ is the angular bisector of the lines $l_{2} : x - y + 1 = 0$ and $l_{3} : α x + β y + 17 = 0$ , then $α^{2} + β^{2} - α - β$ is equal to ______.