Question

If point (x+2,2y-3) lies on the line joining the points (1,-3) and (-4,2). Write a linear relation between x and y.

Answer 1

Answer:

x +2y+1=0

Step-by-step explanation:

Concept:

The equation of line joining $(x_1,y_1)\:and(x_2,y_2)\:is \frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}$

The equation of line joining (1,-3) and(-4, 2) is

$\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1} \\\\\frac{y-(-3)}{2-(-3)}=\frac{x-1}{-4-1} \\\\\frac{y+3}{2+3}=\frac{x-1}{-5} \\\\\frac{y+3}{5}=\frac{x-1}{-5} \\\\\frac{y+3}{1}=\frac{x-1}{-1}$

-y-3=x-1

x+y+2=0

since the point (x+2,2y-3) lies on x+y+2=0,

(x+2)+(2y-3)+2=0

x +2y+1=0 which is the required relation.

Answer 2

Answer:  The answer is $x+2y=-1.$

Step-by-step explanation:  Given that the point (x + 2, 2y - 3) lies on the line joining the points (1, -3) and (-4, 2).

First, we will find the equation of the line passing through (1, -3) and (-4, 1).

The slope of the line will be

$m=\dfrac{2+3}{-4-1}=\dfrac{5}{-5}=-1.$

So, the equation of the line is given by

$y+3=-1(x-1)\\\\\Rightarrow y+3=-x+1\\\\\Rightarrow x+y=-2.$

Now, since the point (x + 2, 2y - 3) lies on the line, so we have

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

which is the required linear relation between 'x' and 'y'.