If the points $(1,2)$ and $(3,4)$ lie on the same side of the straight line $3 x-5 y+a=0$, then $a$ lies in…
If the points $(1,2)$ and $(3,4)$ lie on the same side of the straight line $3 x-5 y+a=0$, then $a$ lies in the set
- $[7,11]$
- $\mathbb{R}-[7,11]$
- $[7, \infty)$
- $(-\infty, 11]$
Solution
Since, the points $(1,2)$ and $(3,4)$ lie on the same side of the line $3 x-5 y+a=0$
$
\begin{array}{lrrlrl}
& \therefore & 3(1)-5(2)+a & \geq 0 & \text { or } & \leq 0 \\
\Rightarrow & a-7 & \geq 0 & \text { or } \leq 0 \\
\Rightarrow & a & \geq 7 & \text { or } & a \leq 7 \\
\text { and } & & 3(3)-5(4)+a & \geq 0 & \text { or } \leq 0 \\
\Rightarrow & a-11 & \geq 0 & \text { or } \leq 0 \\
\Rightarrow & & a & \geq 11 \text { or } & a \leq 11
\end{array}
$
So, common condition is $[7,11]$
Asked in: AP EAMCET 2013
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