Question

The line $y=2x+4$ is shifted 2 units along +y keeping parallel and then +1 unit along +x direction in the same manner, then equation of the line in its new position is
a)$y=2x+6$
b)$y=2x+5$
c)$y=2x+4$
d)None of these

Answer 1

Hint: In this case it is given that the line will be shifted parallel to itself. Therefore, its slope will remain the same after the line is shifted. Thus, we should try to find out the intercept of the new line to obtain the answer to this question.

Complete step-by-step answer:
The general equation of a straight line is given by
$y=mx+c..............(1.1)$
Where m is the slope and c is the intercept.
Comparing to the equation of the line given in the question i.e. $y=2x+4$, we find that m=2 and c=4…….. (1.2)
Now, let $\left( {{x}_{0}},{{y}_{0}} \right)$ be any point on the line. Therefore, x and y satisfy the equation ${{y}_{0}}=2{{x}_{0}}+4...............(1.3)$
When the line is shifted 2 units along +y direction and then +1 unit along +x direction, the point $\left( {{x}_{0}},{{y}_{0}} \right)$ would shift to the point $\left( {{x}_{0}}+1,{{y}_{0}}+2 \right)$ and will lie on the new line……………..(1.4)
When the line is shifted parallel to itself, its slope(m) remains the same and only the intercept c changes. Therefore, using the value of m from equation (1.2), the form of the equation of the new line will be
$y=2x+c$
As the point $\left( {{x}_{0}}+1,{{y}_{0}}+2 \right)$ will lie on the new line, it should satisfy the equation of the new line. Thus,
$\begin{aligned} & {{y}_{0}}+2=2\left( {{x}_{0}}+1 \right)+c \\\ & \Rightarrow 2x+4+2=2x+2+c\text{ (from (1}\text{.3))} \\\ & \Rightarrow c=4 \\\ \end{aligned}$
Therefore, using the value of c as 4, the equation of the new line would be $y=2x+4$.

Note: We should note that when the line is shifted parallel to itself, only the slope remains constant. However, the intercepts of the old and new lines will be different.