Question

A point equidistant from the lines $4x+3y+10=0$,$5x-12y+26=0$ & $7x+24y-50=0$ is

• A. $(1,-1)$
• B. $(1,1)$
• C. $(0,0)$
• D. $(0,1)$

Answer 1

The correct option is C

$(0,0)$

Explanation for correct option:

Step1. Expressing the given data.

Given ,

$4x+3y+10=0...\left(i\right)$

$5x-12y+26=0...\left(ii\right)$

$7x+24y-50=0....\left(iii\right)$

Let $P(x_1,y_1)$ be the point equidistance from the given line equations.

Distance of a point $P(x_1,y_1)$ from a line is given $ax+by+c=0$is

$d=\frac{\left|a{x}_1+b{y}_1+c\right|}{\sqrt{a^2+b^2}}$

Step2. Finding distance of point from all the line

So, the distance of point $P(x_1,y_1)$ from line $1$is

$\begin{array}{rcl}{d}_1& =& \frac{\left|4x_1+3y_1+10\right|}{\sqrt{4^2+3^2}}\\ & =& \frac{\left|4x_1+3y_1+10\right|}{5}...\left(a\right)\end{array}$

Distance from line $2$is

$\begin{array}{rcl}{d}_2& =& \frac{\left|5x_1-12y_1+26\right|}{\sqrt{5^2+{12}^2}}\\ & =& \frac{\left|5x_1-12y_1+26\right|}{13}...\left(b\right)\end{array}$

Distance from line $3$ is

$\begin{array}{rcl}{d}_3& =& \frac{\left|7x_1+24y_1-50\right|}{\sqrt{7^2+{24}^2}}\\ & =& \frac{\left|7x_1+24y_1-50\right|}{25}...\left(c\right)\end{array}$

Equate equation $\left(a\right),\left(b\right)\&\left(c\right)$ as distance of the point from the three line are equal

$\begin{array}{rcl}\frac{\left|4x_1+3y_1+10\right|}{5}& =& \frac{\left|4x_1+3y_1+10\right|}{5}=\frac{\left|7x_1+24y_1-50\right|}{25}\end{array}$

Step3. Checking option

Now check the option.

As, only $x_1=0,y_1=0$satisfy the condition.

Therefore the point is $(0,0)$

Hence, correct option is $\left(C\right)$