Question

If each of the points (x1,4),(-2,y1) lie on the line joining the points (2,-1),(5,-3) then the point P(x1,y1) lies on the line..... ??????

Answer 1

equation of line joining (2,-1),(5,-3) is:

2x + 3y -1 = 0

Now (x1,4) , (-2,y1) lie on 2x + 3y -1 = 0. So it will satisfy.

2x1 + 3*4 -1 = 0

and

2*-2 + 3y1 -1 = 0

x1=-11/2 and y1 = 5/3

Now 2(-11/2) + 3(5/3) -1 does not= 0

So (x1,y1) does not lie on this line.

I hope it helps..

Answer 2

Answer:

No, $P(x_1,y_1)$ does not lie on the given line.

Step-by-step explanation:

Since, the equation of line joining the points (2,-1) and (5,-3) is,

$y-(-1)=\frac{-3-(-1)}{5-2}(x-2)$

$y+1=\frac{-3+1}{3}(x-2)$

$y+1=\frac{-2}{3}(x-2)$

$3y+3=-2x+4$

$2x+3y=1$ -------(1),

Now, we have given,

$(x_1,4)$  and $(-2,y_1)$  lie on the line (1),

⇒ They must satisfy equation (1),

$2x_1+3\times 4=1$

$2x_1+12=1$

$2x_1=-11$

$\implies x_1=-\frac{11}{2}$

$2\times -2+3y_1=1$

$-4+3y_1=1$

$3y_1=5$

$\implies y_1=\frac{5}{3}$

Thus,   $(x_1,y_1)=(-\frac{11}{2},\frac{5}{3})$

Since,

When we put, (-11/2, 5/3 ),

$2\times -\frac{11}{2}+3\times \frac{5}{3}\neq 1$

⇒ (-11/2, 5/3 ) is not satisfying equation (1),

⇒ $(x_1,y_1)$ does not lie on the line (1).