Question

Find the position of the point (8, −9) with respect to the lines $2x+3y-4=0$ and $6x+9y+8=0$
A.Point lies on the same side of the lines
B.Point lies on one of the lines
C. Point lies on the different side of the lines
D. Point lies between the lines.

Answer 1

Hint: Here we will be substituting the point (8, −9) in both the equation of line $2x+3y-4=0$ and $6x+9y+8=0$.You have to also observe that both the given lines are parallel. We will then see what value we got in either case and decide the answer.

Complete step by step solution:

Let L1 be the line $2x+3y-4=0$
Let L2 be the line $6x+9y+8=0$
Let P be the point (8, −9)

∴ ${{L}_{1}}:2x+3y-4=0$, ${{L}_{2}}:6x+9y+8=0$ & P(8, −9)

Substituting point P in L1:
We get, $2\times 8+3x\left( -9 \right)-4=16-27-4=-15<0$

Here, the value is less than zero i.e. negative.
We will now substitute point P in the line L2:
We get,
$\Rightarrow 6\times 8+9x\left( -9 \right)+8$
$\Rightarrow 48-81+8=-25<0$

Here also we got the value as negative i.e. less than zero.

Since after substituting the point P(8, −9) in both the lines L1 and L2, We are getting negative values i.e. sign of both the values (25 & 18) as negative

We can say that point P(8, −9) lies on the same side of the lines.
Hence, the Correct option is A
The point lies on the same side of the lines

Note:
1)If coefficients of x & y in both lines are the same or can be made the same by multiplying a number, then the lines are parallel to each other
2) Let P(x1 y1) be a random point and there be two lines ;
\[{{L}_{1}}:\,a,x+b,\,y+{{c}_{1}}\,=\,0\]
\[{{L}_{2}}:\,{{a}_{2}}x+{{b}_{2y}},\,{{c}_{2}}\,=\,0\]
To check the position of point P(x1 y1) w. r. t lines L1 & L2 we substitute point P in both lines. We conclude that if after substituting signs of the values we got
are same i.e both positive or both negative then point lies on the same side of lines