If the points(1, 2) and (3, 4) were to be on the same side of
the line 3x - 5y + a = 0, then
(a) 7
(b) a = 7
(c) a = 1
(d) a <7 or a > 11
[EAMCET 2000]
Answers
ANSWER:-
3x-5y+a=0
i) (1,2) x=1 , y=2
.°. 3×1 - 5×2 + a=0
.°. 3 - 10 + a=0
.°. a<7
ii) (3,4) x=3 , y=4
.°. 3×3 - 5×4 + a = 0
.°. 9 - 20 + a =0
.°. a>11
so, a<7, or a>11
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The correct answer is a < 7 or a > 11.
Given: Points (1, 2) and (3, 4).
Equation of line = 3x - 5y +a = 0.
To Find: Range of a.
Solution:
Put the point (1, 2) in the equation 3x - 5y +a = 0.
= 3(1) - 5(2) + a
= (a - 7)
Put the point (3, 4) in the equation 3x - 5y +a = 0.
= 3(3) - 5(4) + a
= (a - 11)
So as both the points are in the same side of line both should give same sign when put in equation.
So by multiplying both we will always get positive either both are positive or both are negative.
(a - 7)(a - 11) > 0
a < 7 or a > 11.
Hence, the range of a is a < 7 or a > 11.
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