Question

If the point $(2k-3,k+2)$ lies on the line $2x+3y+15=0,$ find the value of $k$.

• A. $k=\frac{-15}{7}$
• B. $k=\frac{-13}{7}$
• C. $k=\frac{15}{7}$
• D. $k=\frac{13}{7}$

Answer 1

The correct option is A

$k=\frac{-15}{7}$

Explanation for the correct option:

Point will lie on line

The point $(2k-3,k+2)$ lies on the line $2x+3y+15=0,$

This means that the point will satisfy the equation.

So, if we put the value of $x=2k-3$ and $y=k+2$ in the equation $2x+3y+15=0,$ then, we must get the $RHS$ as zero.

$$\begin{gathered} LHS \\ =2(2k-3)+3(k+2)+15 \\ =4k-6+3k+6+15 \\ =7k+15 \end{gathered}$$

Equating $LHS\text{and}RHS$

$$\begin{gathered} \Rightarrow 7k+15=0. \\ \Rightarrow 7k=-15 \\ \Rightarrow k=-\frac{15}{7} \end{gathered}$$

Hence, the value of $k=-\frac{15}{7}$.

Hence, option (A) is the correct answer