
A point P divides the line segment joining the points A (3,-5) and B (-4,8) such that $\dfrac{AP}{PB}=\dfrac{k}{1}$ . If P lies on the line $x+y=0$ , then find the value of k.
Answer
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Hint: We will use the section formula, $\left( a,b \right)=\left( \dfrac{{{m}_{1}}{{x}_{2}}+{{m}_{2}}{{x}_{1}}}{{{m}_{1}}+{{m}_{2}}},\dfrac{{{m}_{1}}{{y}_{2}}+{{m}_{2}}{{y}_{1}}}{{{m}_{1}}+{{m}_{2}}} \right)$ where a point $\left( a,b \right)$ divides a line with end coordinates $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ in the ratio ${{m}_{1}}:{{m}_{2}}$internally, to find the coordinate of point P. The values of point P will contain the variable k. Then we will substitute this value of P in the equation $x+y=0$, to find the value of k.
Complete step by step answer:
Here the ratio is given as $\dfrac{AP}{PB}=\dfrac{k}{1}$ which clearly implies that the point P divides the line segment joining the points A and B internally.
We know if a point $\left( a,b \right)$ divides a line with end coordinates $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ in the ratio ${{m}_{1}}:{{m}_{2}}$internally, then the coordinates of point (a,b) is given by
$\left( a,b \right)=\left( \dfrac{{{m}_{1}}{{x}_{2}}+{{m}_{2}}{{x}_{1}}}{{{m}_{1}}+{{m}_{2}}},\dfrac{{{m}_{1}}{{y}_{2}}+{{m}_{2}}{{y}_{1}}}{{{m}_{1}}+{{m}_{2}}} \right)\text{ }\ldots \left( i \right)$
If a point P (a,b) divides the line segment with coordinates A (3,-5) and B (-4,8) in the ratio k:1, then substituting these values in equation (i), we get
$\begin{align}
& \left( a,b \right)=\left( \dfrac{k(-4)+1\cdot 3}{k+1},\dfrac{k\cdot 8+1(-5)}{k+1} \right) \\
& \left( a,b \right)=\left( \dfrac{-4k+3}{k+1},\dfrac{8k-5}{k+1} \right)
\end{align}$
It is given that the point P lies on the line $x+y=0$
This means that the coordinates of point P satisfy the equation $x+y=0$
Substituting the value of (a,b) in this equation, we get
$\begin{align}
& \text{ }\dfrac{-4k+3}{k+1}+\dfrac{8k-5}{k+1}=0 \\
& \Rightarrow \dfrac{-4k+3+8k-5}{k+1}=0 \\
& \Rightarrow 4k-2=0 \\
& \Rightarrow 4k=2 \\
& \Rightarrow k=\dfrac{1}{2} \\
\end{align}$
So, the value of k is $\dfrac{1}{2}$.
Note: We should keep a cool mind while doing calculations to make it error free. We can use the formula $\left( a,b \right)=\left( \dfrac{{{x}_{1}}+k{{x}_{2}}}{k+1},\dfrac{{{y}_{1}}+k{{y}_{2}}}{k+1} \right)$ directly when (a,b) divides a line with end coordinates $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ in the ratio k:1.
Complete step by step answer:
Here the ratio is given as $\dfrac{AP}{PB}=\dfrac{k}{1}$ which clearly implies that the point P divides the line segment joining the points A and B internally.
We know if a point $\left( a,b \right)$ divides a line with end coordinates $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ in the ratio ${{m}_{1}}:{{m}_{2}}$internally, then the coordinates of point (a,b) is given by
$\left( a,b \right)=\left( \dfrac{{{m}_{1}}{{x}_{2}}+{{m}_{2}}{{x}_{1}}}{{{m}_{1}}+{{m}_{2}}},\dfrac{{{m}_{1}}{{y}_{2}}+{{m}_{2}}{{y}_{1}}}{{{m}_{1}}+{{m}_{2}}} \right)\text{ }\ldots \left( i \right)$
If a point P (a,b) divides the line segment with coordinates A (3,-5) and B (-4,8) in the ratio k:1, then substituting these values in equation (i), we get
$\begin{align}
& \left( a,b \right)=\left( \dfrac{k(-4)+1\cdot 3}{k+1},\dfrac{k\cdot 8+1(-5)}{k+1} \right) \\
& \left( a,b \right)=\left( \dfrac{-4k+3}{k+1},\dfrac{8k-5}{k+1} \right)
\end{align}$
It is given that the point P lies on the line $x+y=0$
This means that the coordinates of point P satisfy the equation $x+y=0$
Substituting the value of (a,b) in this equation, we get
$\begin{align}
& \text{ }\dfrac{-4k+3}{k+1}+\dfrac{8k-5}{k+1}=0 \\
& \Rightarrow \dfrac{-4k+3+8k-5}{k+1}=0 \\
& \Rightarrow 4k-2=0 \\
& \Rightarrow 4k=2 \\
& \Rightarrow k=\dfrac{1}{2} \\
\end{align}$
So, the value of k is $\dfrac{1}{2}$.
Note: We should keep a cool mind while doing calculations to make it error free. We can use the formula $\left( a,b \right)=\left( \dfrac{{{x}_{1}}+k{{x}_{2}}}{k+1},\dfrac{{{y}_{1}}+k{{y}_{2}}}{k+1} \right)$ directly when (a,b) divides a line with end coordinates $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ in the ratio k:1.
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