Question

The line passing through the points (a, b) and (-a,-b) also passes through which of the following points?

Answer 1

Hint: In this type of problem, we find the equation of the line first through the given two points and we substitute the options given in the lines to check whether points lie on the lines or not.

Complete step by step Answer:

Consider the points P$\left( a,\ b \right)$ and Q$\left( -a,-b \right)$We use this formula:
If two points \[\left( {{x}_{1}}\ \ {{y}_{1}} \right)\] and \[\left( {{x}_{2}}\ \ {{y}_{2}} \right)\]are given; equation of line passing through these points are:

\[y-y,=\dfrac{\left( {{y}_{2}}-{{y}_{1}} \right)}{\left( {{x}_{2}}-{{x}_{1}} \right)}\left( x-{{x}_{1}} \right)\]
OR, \[\left( y-{{y}_{1}} \right)\left( {{x}_{2}}-{{x}_{1}} \right)=\left( x-{{x}_{1}} \right)\left( {{y}_{2}}-{{y}_{1}} \right)\]
Now, we substitute points in the formula to find equation of line:

 Equation of line through P$\left( a,\ b \right)$& Q$\left( -a,-b \right)$

\[\Rightarrow \left( y-b \right)\left( -a-a \right)=\left( x-a \right)\left( -b-b \right)\]
\[\Rightarrow \left( y-b \right)\left( -2a \right)=\left( x-a \right)\left( -2b \right)\]
Dividing by -2 on both sides we get,

\[\Rightarrow \left( y-b \right)a=\left( x-a \right)b\]


Multiplying with the terms in the bracket we get,

\[\Rightarrow ay-ab=b-ab\]

Cancelling -ab on both sides we get,

\[\Rightarrow ay=bx\]

Dividing by ab on both sides we get,

\[\Rightarrow \dfrac{x}{a}=\dfrac{y}{b}\]
\[\Rightarrow \dfrac{x}{a}-\dfrac{y}{b}=0\]
 Equation of line passing through points PQ is $\dfrac{x}{a}-\dfrac{y}{5}=0$

Now, we will substitute points in the above equation to check if the point given in the options lies on the line. Substituting points in the line equation we get,

A. $\left( 1,\ 1 \right)\to \dfrac{1}{a}=\dfrac{1}{b}\ne 0\to $Does not lies on the line
B. $\left( 3a,-2b \right)\to \dfrac{3a}{a}=\dfrac{\left( -2b \right)}{b}=5\ne 0\to $ Does not lie on the line
C. $\left( {{a}^{2}},\ ab \right)\to \dfrac{{{a}^{2}}}{a}=\dfrac{ab}{b}=0\to $ Hence, Lies on the line
D. $\left( 4a,-2b \right)\to \dfrac{4a}{a}=\dfrac{2b}{b}\ne 0\to $Does not lie on line.

The correct option is C (a², ab)

Note: We got the line equation as \[y-y,=\dfrac{\left( {{y}_{2}}-{{y}_{1}} \right)}{\left( {{x}_{2}}-{{x}_{1}} \right)}\left( x-{{x}_{1}} \right)\] . Here, if the denominator part is equal to zero, then the result won’t be infinity but instead we will get a line parallel to the y-axis i.e. line equation will be of the form x=constant. You must also be very careful in substituting the points in the line equation. This equation of line can also be written as:\[\left( y-{{y}_{1}} \right)\left( {{x}_{2}}-{{x}_{1}} \right)=\left( x-{{x}_{1}} \right)\left( {{y}_{2}}-{{y}_{1}} \right)\]. You can see that if the denominator term $\left( {{x}_{2}}-{{x}_{1}} \right)$ is equal to zero, we will get the line of the form $x={{x}_{1}}$ which is a line parallel to y axis.