Question

A straight line passing through the point (x1,y1) meets the positive coordinate at a,b.
The locus of the point p which divides ab in the ratio l:m is​

Answer 1

SOLUTION

GIVEN

A straight line passing through the point $\sf{(x_1,y_1)}$ meets the positive coordinate at A , B

TO DETERMINE

The locus of the point p which divides AB in the ratio l : m

EVALUATION

Here it is given that a straight line passing through the point $\sf{(x_1,y_1)}$ meets the positive coordinate at A , B

Let the equation of the line is

$\displaystyle \sf{ \frac{x}{a} + \frac{y}{b} = 1 \: \: \: - - - (1) }$

The line cuts x axis at A(a, 0) & y axis at B(0,b)

Since the line goes through the point $\sf{(x_1,y_1)}$

Thus we have

$\displaystyle \sf{ \frac{x_1}{a} + \frac{y_1}{b} = 1 \: \: \: - - - (2) }$

Let the coordinates of the point P is (h, k)

Now the point P divides the line AB in the ratio l : m

Thus we get

$\displaystyle \sf{ h = \frac{ma}{l + m} \: \: \: and \: \: k = \frac{lb}{l + m} }$

$\displaystyle \sf{ \implies \: a = \frac{(l + m)h}{ m} \: \: \: and \: \: \: b= \frac{(l + m)k}{l } }$

Putting the values of a and b in Equation 2 we get

$\displaystyle \sf{ \frac{x_1}{\frac{(l + m)h}{ m}} + \frac{y_1}{\frac{(l + m)k}{ l}} = 1 }$

$\displaystyle \sf{ \implies \: \frac{mx_1}{(l + m)h} + \frac{ly_1}{(l + m)k} = 1 }$

$\displaystyle \sf{ \implies \: \frac{mx_1}{h} + \frac{ly_1}{k} = l + m}$

Hence the locus of the point P is

$\displaystyle \sf{ \implies \: \frac{mx_1}{x} + \frac{ly_1}{y} = l + m}$

FINAL ANSWER

Hence the required locus of the point P is

$\boxed{ \: \: \displaystyle \sf{ \frac{mx_1}{x} + \frac{ly_1}{y} = l + m} \: \: \: }$

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