A straight line passes through P(1, 2) such t | vedclass

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(D) Let the equation of the line in intercept form be $\frac{x}{a} + \frac{y}{b} = 1$.The intercepts on the $x$-axis and $y$-axis are $A(a, 0)$ and $B

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$2x + y - 4 = 0$

Vedclass · Answer

(D) Let the equation of the line in intercept form be $\frac{x}{a} + \frac{y}{b} = 1$.The intercepts on the $x$-axis and $y$-axis are $A(a, 0)$ and $B(0, b)$ respectively.The point $P(1, 2)$ is the midpoint of the segment $AB$.Using the midpoint formula,we have $\frac{a + 0}{2} = 1$ and $\frac{0 + b}{2} = 2$.This gives $a = 2$ and $b = 4$.Substituting these values into the intercept form equation,we get $\frac{x}{2} + \frac{y}{4} = 1$.Multiplying by $4$,we get $2x + y = 4$,or $2x + y - 4 = 0$.

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

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