Double Intercept Form

Here, we will discuss about the double intercept form of the equation of straight lines.

Let a straight line intersect x-axis and y-axis at $A(a,0)$ and $B(0,b)$ respectively. Then, the intercepts on the x-axis and y-axis are $OA=a$ and $OB=b$ respectively. Also, let $P(x,y)$ be any point on $AB$.

Then, $$\text{Slope of AP}=\text{Slope of PB}$$ $$\frac{y-0}{x-a}=\frac{b-y}{0-x}$$ $$\frac{y}{x-a}=\frac{b-y}{-x}$$ $$-xy=bx-xy-ab+ay$$ $$ab=bx+ay$$ $$1=\frac{x}{a}+\frac{y}{b}$$ $$\therefore \frac{x}{a}+\frac{y}{b}=1$$

This is the equation of the line whose intercepts are $a$ and $b$ and is known as the double intercept form. This form of the equation of a straight line cannot be used if the line passes through the origin or if it be parallel to any one of the axes.

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Obtain the equation of the straight line passing through the point $(3,4)$ cutting off equal intercepts on the axes.

Here, $$\text{x-intercept}=\text{y-intercept}$$ $$\therefore a=b$$ Then the equation of the line is, $$\frac{x}{a}+\frac{y}{b}=1$$ $$\frac{x}{a}+\frac{y}{a}=1$$ $$x+y=a\text{ \_\_(1)}$$

Since this line passes through the point $(3,4)$, $$3+4=a$$ $$\therefore a=7$$

Putting the value of $a$ in $\text{(1)}$, $$x+y=7$$ which is the required equation of the line.

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Find the equation of the straight line which passes through the point $(3,4)$ and makes intercepts on the axes, the sum of whose lengths is $14$.

$$\text{Let x-intercept}=a$$ $$\text{y-intercept}=b$$ $$\text{Then, }a+b=14$$ $$b=14-a\text{ \_\_(1)}$$

The equation of the line is, $$\frac{x}{a}+\frac{y}{b}=1$$ $$bx+ay=ab$$ Since this line passes through $(3,4)$, $$3b+4a=ab\text{ \_\_(2)}$$

Solving $\text{(1)}$ and $\text{(2)}$, $$3(14-a)+4a=a(14-a)$$ $$42-3a+4a=14a-a^2$$ $$a^2-13a+42=0$$ $$(a-6)(a-7)=0$$ $$\therefore a=6,7$$

Taking $a=6$, $b=14-6=8$. Then, the required equation of the line is $$\frac{x}{6}+\frac{y}{8}=1$$ $$\therefore 4x+3y=24$$

Taking $a=7$, $b=14-7=7$. Then, the required equation of the line is, $$\frac{x}{7}+\frac{y}{7}=1$$ $$\therefore x+y=7$$

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Determine the equation of the line the portion of which, intercepted by the axes, is divided by the point $(-5,4)$ in the ratio $1:2$.

Let the line intersect x-axis and y-axis at $A(a,0)$ and $B(0,b)$ respectively and $AB$ is divided at $C$ in the ratio $1:2$. Then, from section formula for internal division, we have, $$-5=\frac{1\times 0+2\times a}{1+2}\text{ and }4=\frac{1\times b+2\times 0}{1+2}$$ $$-5=\frac{2a}{3}\text{ and }4=\frac{b}{3}$$ $$\therefore a=\frac{-15}{2}\text{ and }b=12$$

Thus, equation of line is, $$\frac{x}{a}+\frac{y}{b}=1$$ $$\frac{-2x}{15}+\frac{y}{12}=1$$ $$-8x-5y=60$$ $$\therefore 8x-5y+60=0$$

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