Question

A straight line through $P(1,2)$ is such that its intercept between the axes is bisected at $P$. Its equation is

• A. $x+2y=5$
• B. $x–y+1=0$
• C. $x+y–3=0$
• D. $2x+y–4=0$

Answer 1

The correct option is D

$2x+y–4=0$

Explanation for the correct answer:

Step 1: Finding the point A and B.

The straight line passes through the point $P(1,2)$

The line touches the $X-$axis at $A(a,0)$ and $Y-$axis at $B(0,b)$

The mid point at $X-$axis is

$$\begin{gathered} \frac{a+0}{2}=1 \\ a=2 \end{gathered}$$

The mid point at $Y-$axis is

$$\begin{gathered} \frac{0+b}{2}=2 \\ b=4 \end{gathered}$$

Step 2: Finding the slope.

The slope of the line $A(2,0)$ and $B(0,4)$ is

$$\begin{gathered} m=\frac{y_2-y_1}{x_2-x_1} \\ =\frac{4-0}{0-2} \\ =-2 \end{gathered}$$

Step 3: Finding the equation of the line.

The equation of the line passing through $(1,2)$ with the slope $m=-2$ is

$$\begin{gathered} \Rightarrow (y-y_1)=m(x-x_1) \\ (y-2)=-2(x-1) \\ y-2=-2x+2 \\ 2x+y-4=0 \end{gathered}$$

Hence, option (D) $2x+y-4=0$ is the correct answer.