Question

Find the equation of the line passing through the point (7,9)and which is such that the portion of it between the axes is divided by the point in the ratio 3:1

Answer 1

Answer:

The equation of the required line is 3x+7y-84=0

Step-by-step explanation:

$\text{Let the required line meets the coordinate axes at A(a,0) and B(0,b)}$

$\text{As per given data, the point (7,9) divides AB in the ratio 3:1}$

$\bf(\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n})=(7,9)$

$\implies(\frac{3(0)+1a}{3+1},\frac{3(b)+1(0)}{3+1})=(7,9)$

$\implies(\frac{a}{4},\frac{3b}{4})=(7,9)$

$\implies\frac{a}{4}=7\:\:and\:\:\frac{3b}{4}=9$

$\implies\;a=28\:\:and\:\:\frac{b}{4}=3$

$\implies\;a=28\:\:and\:\:b=12$

$\therefore\text{Equation of the required line is }$

$\bf\frac{x}{a}+\frac{y}{b}=1$

$\frac{x}{28}+\frac{y}{12}=1$

$\frac{3x+7y}{84}=1$

$\implies\boxed{\bf\;3x+7y-84=0}$

Answer 2

The equation of line is 3x+7y=84

Step-by-step explanation:

Let the intercept form of equation of line, $\dfrac{x}{a}+\dfrac{y}{b}=1$

• x-intercept: (a,0)

• y-intercept: (0,b)

The portion of line between the axes is divided by the point (7,9) in the ratio 3:1

Using section formula

$x=\dfrac{mx_2+nx_1}{m+n}$

$x=7,m=3,n=1,x_1=a,x_2=0$

$7=\dfrac{3\cdot 0+1\cdot a}{3+1}$

$a=28$

$y=\dfrac{my_2+ny_1}{m+n}$

$y=9,m=3,n=1,y_0=a,y_2=b$

$9=\dfrac{3\cdot b+1\cdot 0}{3+1}$

$b=12$

$\dfrac{x}{28}+\dfrac{y}{12}=1$

$3x+7y=84$

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Equation of line

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