Find the equation of the straight line passing through the origin and bisecting the portion of the line ax + by + c = 0 intercepted between the coordinate axes.

Open in App

Solution

$The equation of the line passing through the origin is y = mx$

$Let the line a x + b y + c = 0 meet the coordinate axes at A and B.$

$So, the coordinates of A and B are A$

$(- \frac{c}{a}, 0) a n d B (0, - \frac{c}{b}) .$

$Now, the midpoint of AB is$

$(- \frac{c}{2 a}, - \frac{c}{2 b}) .$

$Clearly, (- \frac{c}{2 a}, - \frac{c}{2 b}) lies on the line y = mx$

$∴ - \frac{c}{2 b} = m \times \frac{- c}{2 a}$

$\Rightarrow m = \frac{a}{b}$

$Hence, the equation of the required line is y = \frac{a}{b} x$

$\Rightarrow a x - b y = 0$

$P.Q. The area of the triangle formed by the coordinate axes and a line is 6 square units and the length of the hypotenues is 5 units. Find the equation of the line.$

$The equation of the line with intercepts$

$a a n d b i s \frac{x}{a} + \frac{y}{b} = 1$

$The line intersects the axes at A (a, 0) and B (0, b)$

$Let O be the origin$

$It is given that the area of the triangle formed by the coordinate axes and the line is 6 square units.$

$∴ A r e a o f Δ O A B = \frac{1}{2} | 0 A \times O B |$

$\Rightarrow 6 = \frac{1}{2} . | a b |$

$\Rightarrow | a b | = 12$

$\Rightarrow | a | b | = 3 \times 4 . . . (i)$

$Here, the length of the hypotenuse is 5 units, i.e., AB = 5$

$∴ O A^{2} + O B^{2} = A B^{2}$

$\Rightarrow a^{2} + b^{2} = 25 . . . (i i)$

$Let us find the possible values of a and b from (i) and (ii).$

$Using equation (i)$

$\frac{144}{b^{2}} + b^{2} = 25$

$\Rightarrow b^{4} - 25 b^{2} + 144 = 0$

$\Rightarrow b^{2} = 16, 9$

$\Rightarrow b \pm 4, \pm 3$

$Now, from equation (i)$

$F o r b = \pm 4, | a | = 3 \Rightarrow b = 4, a = 3,$

$- 3 a n d b = - 4, a = 3, - 3$

$F o r b = \pm 3, | a | = 4 \Rightarrow b = 3, a = 4,$

$- 4 a n d b = - 3, a = 4, 4$

$Hence, the possible equations of the lines are$

$\frac{x}{3} + \frac{y}{4} = 1, \frac{x}{- 3} + \frac{y}{4} = 1, \frac{x}{3} + \frac{y}{- 4}$

$= 1, \frac{x}{- 3} + \frac{y}{- 4} = 1, \frac{x}{4} + \frac{y}{3} = 1, \frac{x}{4} +$

$\frac{y}{3} = 1, \frac{x}{4} + \frac{y}{3} = 1 a n d \frac{x}{- 4} + \frac{y}{3} = 1$

$o r 4 x + 3 y = 12 o r 4 x - 3 y = - 12 o r$

$- 4 x + 3 y = - 12 o r - 4 x - 3 y = 12 o r 3 x$

$+ 4 y = 12 o r x - 4 y = - 12 o r - 3 x + 4 y$

$= - 12 o r - 3 x - 4 y = 12$

Suggest Corrections

Similar questions

Q. Find the equation of the straight line which passes through

(2, 3)

and the portion of the line intercepted between the axes is bisected at this point.

Q. If a straight line passing through the point

P (- 3, 4)

is such that its intercepted portion between the coordinate axes is bisected at

P

, the its equation is :

Q. Find the equations of the straight lines which pass through the origin and trisect the portion of the straight line

2 x + 3 y = 6

which is intercepted between the axes.

Find the equation of the straight line passing through the point (2, 1) and bisecting the portion of the straight line 3x - 5y = 15 lying between the axes.

Q. A straight line passes through the point (α, β) and this point bisects the portion of the line intercepted between the axes. Show that the equation of the straight line is

\frac{x}{2 α} + \frac{y}{2 β} = 1

Join BYJU'S Learning Program