Question

The area of the triangle formed by the lines is $7x-2y+10=0,$ $7x+2y-10=0$ and $y+2=0$ is

• A. $8sq.unit$
• B. $12sq.unit$
• C. $14sq.unit$
• D. none of these

Answer 1

The correct option is C $14sq.unit$

Given,

$L_1\to 7x-2y+10=0$

$L_2\to 7x+2y-10=0$

$L_3\to y+2=0$

on solving $L_1$ & $L_2$ we can find a point of intersection of these lines.

$7x-2y+10=0$

$7x+2y+10=0$

Add them, $14x=0$

$x=0$

put the value of $x$ is $L_2\to 7\left(0\right)+2y-10=0$

$y=5$

$\therefore P.O.I$ of $L_1$ & $L_2$ is $P(0,5)$

Now,

$L_1$ & $L_3$

$7x-2y+10=0$

$y+2=0$

from $L_{3}y=-2$

put in $L_1,7x-2(-2)+10=0$

$x=-2$

$Q(-2,-2)$

Now,

$L_2$ & $L_3$

$7x+2y-10=0$

$y+2=0$

from $L_{3}y=-2$

put in $L_2,7x+2(-2)-10=0$

$x=2$

$R(2,-2)$

Therefore, Area of $△$ formed by joining $P,Q,R$ is Area of a trio

$=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$

$x_1=0,x_2=-2,x_3=2,y_1=5,y_2=-2,y_3=-2$

Area $=\frac{1}{2}|o(-2-(-2\left)\right)+(-2)\left[\right(-2)-5]+2[5-(-2\left)\right]$

Area $=\frac{1}{2}|0+14+14|$

Area $=14uni{t}^2$