Question

The point \[\text{P}\left( a,b \right)\] lies on the straight line \[3x+2y\ =\ 13\] and the point \[\text{Q}\left( b,a \right)\] lies on the straight line \[4x-y\ =\ 5\], then the equation of line PQ is
A. \[x-y\ =\ 5\]
B. \[x+y\ =\ -5\]
C. \[x+y\ =\ 5\]
D. \[x-y\ =\ -5\]

Answer 1

Hint: First substitute \[\left( a,b \right)\] in \[3x+2y\ =\ 13\] and \[\left( b,a \right)\] in \[4x-y\ =\ 5\], then solve simultaneously to find value of a, b respectively then find the value of points of \[\left( a,b \right)\] and \[\left( b,a \right)\] then find slope and hence equation of line.

Complete Step-by-Step solution:
We are given a point O such that it is given as \[\left( a,b \right)\] and lies on \[3x+2y\ =\ 13\] and we are given another point Q such that it is given as \[\left( b,a \right)\] and lies on \[4x-y\ =\ 5\], then we have to find equation of line.
Now, we know that P is a point with coordinates \[\left( a,b \right)\] which lies on \[3x+2y\ =\ 13\] which means it satisfies for value \[x\ =\ a\] and \[y\ =\ b\].
So, we can write the equation \[3x+2y\ =\ 13\] as \[3a+2b\ =\ 13\] …………………………..(i)
Now, for Q as we know its coordinates as \[\left( b,a \right)\] which lies on straight line \[4x-y\ =\ 5\] which means it satisfy for values \[x\ =\ b\] and \[y\ =\ a\].
So, we can write the equation \[4x-y\ =\ 5\] as \[4b-a\ =\ 5\] ……………………………………(ii)
Now, we will first consider equation (ii) such as,
\[4b-a\ =\ 5\]
Which can be written as,
\[4b\ =\ 5+a\]
So, \[a\ =\ 4b-5\]
Now, will substitute the value of \[a\ =\ 4b-5\] in equation (i) so we get,
\[3a+2b\ =\ 13\]
\[3\left( 4b-5 \right)+2b\ =\ 13\]
On simplifying we get,
\[12b-15+2b\ =\ 13\]
We can write it as,
\[14b\ =\ 28\]
Hence, the value of \[b\ =\ 2\].
Now, as we know the value of \[b\ =\ 2\] so, we can find the value of a by using the fact \[a\ =\ 4b-5\].
So, \[a\ =\ 4\times 2-5\ =\ 3\].
Hence, the value of \[a\ =\ 3\].
So, the point \[\text{P}\left( a,b \right)\] or \[\text{P}\left( 3,2 \right)\] and \[\text{Q}\left( b,a \right)\] or \[\text{Q}\left( 2,3 \right)\].
We know the points \[\text{P}\left( 3,2 \right)\] and \[\text{Q}\left( 2,3 \right)\] so we find the slope by using the formula
\[m\ =\ \dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\] if points are \[\left( {{x}_{1}},{{y}_{1}} \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)\]
So, the slope is \[m\ =\ \dfrac{3-2}{2-3}\ =\ -1\].
Now, we know the slope of line and also one of the points so we can find the equation using formula,
\[y-{{y}_{1}}\ =\ m\ \left( x-{{x}_{1}} \right)\]
Where \[\left( {{x}_{1}},{{y}_{1}} \right)\] is point and m is slope.
Here slope is -1 and point is \[\left( 3,2 \right)\]. So, equation is,
\[y-2\ =\ -1\left( x-3 \right)\]
On simplification we can write it as,
\[y-2\ =\ -x+3\]
Which can be written as,
\[x+y\ =\ 5\]
Hence, the correct option is ‘b’.

Note: One can also find the equation by using three points \[\left( 2,3 \right)\] which will also fetch the same results. So, we can do it as for \[\left( 2,3 \right)\] we get,
 \[y-3\ =\ -1\left( x-2 \right)\]
\[y-3\ =\ -x+2\]
Hence, \[x+y\ =\ 5\] which is the same as above.