Question

Two points $(a,0)$ and $(0,b)$ are joined by a straight line. Another point on this line is $(1,\frac{(a-m)}{a}b)$. find $m$.

• A. $2$
• B. $1$
• C. $0$
• D. $-1$

Answer 1

Answer: (B)

$1$

Slope of a line joining points $(x_1,y_1)\&(x_2,y_2)=\frac{y_2-y_1}{x_2-x_1}$

$\therefore$ Slope of line joining $(a,0)\&(0,b)=\frac{b-0}{0-a}=-\frac{b}{a}$

The equation of line passing through the point $(x_1,y_1)\&(x_2,y_2$ is-

$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

$\therefore$ Equation of line passing through $(a,0)\&(0,b)$ is-

$y-0=\frac{b-0}{0-a}\times (x-a)$

$\Rightarrow y=-\frac{b}{a}x+(-\frac{b}{a}\times -a)$

$\Rightarrow y=-\frac{b}{a}x+b$

As the point $(1,\frac{(a-m)}{a}b)$ lies on the same line, it must satisfies the equation of line.

$\therefore \frac{(a-m)}{a}b=-\frac{b}{a}\times 1+b$

$\Rightarrow \frac{(a-m)}{a}b=\frac{b}{a}(-1+a)$

$\Rightarrow a-m=a-1$

$\Rightarrow m=1$

Hence, $m=1$.