Two points are given (-2,0) and (0,4/3) we can easily find the equation of straight line joining these two points
Let A(-2,0) and B(0,4/3)
Equation of straight line joining these two points $(x_1,y_1) $ and $(x_2,y_2)$ is given by $y-y_1 = \frac{y_2-y_1}{x_2-x_1}(x-x_1)$
Therefore the equation is $2x-3y = -4$
But how can we find that any point joining the above two points if of the form $((-2+2k, 4k/3)$ Please guide on this will be of great help. Thanks
On
Hint if the equation which you have derived is satisfied by $(-2+2k,4k/3)$ then obviously any point lying on that is of that form and yes it satisfies thus proved
On
Hint:
Think about the parametric equation of a line:
$$(x,y)=(x_1,y_1)+k(a,b)$$
Where $(x_1,y_1)$ is a point of the line and $(a,b)$ is the direction of the line, i.e. vector $\stackrel{\rightarrow}{AB}$.
On
Notice, there are infinite points lying on a given line: $2x-3y=-4$ hence in order to find a general/parametric point on the given line, one should find the value of one variable in terms of other as follows $$2x-3y=-4\implies y=\frac{2x+4}{3}$$ If we assume $x=k$ then corresponding y-coordinate $$y=\frac{2k+4}{3}$$ Thus, a general point on the given line is also $\left(k, \frac{2k+4}{3}\right)$ which duly satisfy the equation of the given line: $2x-3y=-4$
If you take other point $\left(-2+2k, \frac{4k}{3}\right)$ this point also satisfies the equation of the given line: $2x-3y=-4$
Thus, one find different forms of general point but they all must satisfy the equation of the given line.
On
Consider the two position vectors of $A$ and $B$:
$$\vec{a} = \left (-2, 0 \right)$$ $$\vec{b} = \left (0, {4\over 3} \right)$$
Now, every point on the line through $AB$ can be described as a vector to $A$ to which a multiple of the vector $\vec{b}-\vec{a}$ is added:
$$\vec{p} = \vec{a} + t(\vec{b}-\vec{a})$$
Thus:
$$\vec{p} = \left (-2, 0 \right) + t*\left (2, {4\over 3} \right) = \left (-2, 0 \right) + \left (2t, {4\over 3}t \right) = \left (2t - 2, {4\over 3}t \right)$$
If you already know what form the points will take and you are merely interested in verifying that the form is indeed correct, then you can just plug the values in and make sure they satisfy the equation.
$$2x - 3y = 2(-2+2k)-3\frac{4k}{3} = -4 + 4k - 4k = 4 $$
so indeed, we conclude $2x - 3y = 4$ is satisfied.