Question

If each of the points $\left(a,6\right)$ and $\left(3,b\right)$ lies on the line joining the point $\left(3,2\right)$ and $\left(5,1\right)$ then the points $\left(a,b\right)$ lies on the line

• A. $3x+4y+7=0$
• B. $2x+3y-7=0$
• C. $4x-3y-7=0$
• D. $3x-2y-7=0$

Answer 1

The correct option is A $3x+4y+7=0$

Line joining $(3,2)$ & $(5,1)$ is

$y-2=\frac{1-2}{5-3}(x-3)$

$2y-4=3-x$

$2y+x=7$ ......(1)

then

$12+a=7$ ......... As $(a,6)$ lies on eqn (1)

$a=-5$

also $2b+3=7$ ......... As $(3,b)$ lies on eqn (1)

$b=2$

then $3x+4y+7=0$

will satisfy $(a,b)$ or $(-5,2)$

$A$ is correct