Question

The ratio by which the line $2x+5y-7=0$ divides the straight line joining the points $(-4,7)$ and $(6,-5)$ is

• A. $1:4$
• B. $1:2$
• C. $1:1$
• D. $2:3$
• E. $1:3$

Answer 1

Answer: (C)

$1:1$

Using the section formula, if a point $(x,y)$ divides the line joining the points $(x_1,y_1)$ and $(x_2,y_2)$ in the ratio $m:n$, then $(x,y)=(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n})$

Let the line $2x+5y-7=0$ divide the straight line joining points $(-4,7)$ and $(6,-5)$ in the ratio $m:n$, then the point of intersection by Section Formula is $(\frac{-4m+6n}{m+n},\frac{7m-5n}{m+n})$

Since the intersection point lies on the given line, we have $2\times \left(\frac{-4m+6n}{m+n}\right)+5\times \left(\frac{7m-5n}{m+n}\right)-7=0$

$\Rightarrow -8m+12n+35m-25n-7m-7n=0$

$\Rightarrow 20m=20n$

$\Rightarrow m:n=1:1$