1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# If a and b are two arbitrary constants, then the straight line (a-2b)x + (a+3b)y + 3a+4b = 0 will pass through

A

(-1,-2)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B

(1,2)

No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
C

(-2,-3)

No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
D

(2,3)

No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
Open in App
Solution

## The correct option is A (-1,-2) Given equation is (a-2b)x + (a+3b)y + 3a+4b = 0 Or a(x+y+3) + b(-2x+3y+4) = 0 This represents a family of straight lines through the point of intersection of x + y + 3 = 0 -----------------------------( 1 ) And -2x + 3y + 4 = 0 ------------------- ( 2 ) Multiplying 2 in equation 1 and add it to equation 2 2x + 2y + 6 = 0 −2x+3y+4=05y+10=0 y = -2 substituting y in equation 1 x - 2 + 3 = 0 x = -1 Point of intersection of these two lines (-1,-2) Family of straight lines should pass through (-1,-2).

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Family of Straight Lines
MATHEMATICS
Watch in App
Explore more
Join BYJU'S Learning Program