First slide

Equation of line in Straight lines

Question

The distance between $(1, 1)$ and the point of intersection of two lines $2 x + 3 y + c = 0$ and $3 x + 2 y + c = 0$ is less than $2 \sqrt{2}$ then

Moderate

A

$(- 15, 5)$
B

$(- 10, 0)$
C

$(- 10, 10)$
D

$(- \frac{19}{5}, 0)$

Solution

To get the point of intersection of two lines, eliminate x
$\begin{array}{l} 3 (2 x + 3 y + c) - 2 (3 x + 2 y + c) = 0 \\ 6 x + 9 y + 3 c - 6 x - 4 y - 2 c = 0 \\ 5 y + c = 0 \\ y = - \frac{c}{5} \end{array}$
Substitute $y = - \frac{c}{5}$ in the equation $2 x + 3 y + c = 0$
$\begin{array}{l} 2 x + 3 (- \frac{c}{5}) + c = 0 \\ 10 x - 3 c + 5 c = 0 \\ 10 x + 2 c = 0 \\ x = - \frac{c}{5} \end{array}$
Hence, the point of intersection is $(- \frac{c}{5}, - \frac{c}{5})$
Given that the distance between $(1, 1)$ and $(- \frac{c}{5}, - \frac{c}{5})$ is $\sqrt{{(1 + \frac{c}{5})}^{2} + {(1 + \frac{c}{5})}^{2}} < 2 \sqrt{2}$
It implies that $\sqrt{2 {(1 + \frac{c}{5})}^{2}} < 2 \sqrt{2}$
Squaring and then simplify
$\begin{matrix} 2 {(1 + \frac{c}{5})}^{2} < 8 \\ {(1 + \frac{c}{5})}^{2} < 4 \end{matrix}$
Apply square root on both sides
$\begin{matrix} - 2 < 1 + \frac{c}{5} < 2 \\ - 3 < \frac{c}{5} < 1 \\ - 15 < c < 5 \end{matrix}$
Therefore, $c \in (- 15, 5)$

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