Question

Find the distance of the point \[\left( {2,3} \right)\;\] form the line \[2x - 3y + 9 = 0\;\] measurement along a line \[x - y + 1 = 0\] .
A.$\dfrac{1}{{\sqrt 2 }}$
B.$4\sqrt 2 $
C.$\sqrt 2 $
D.$\dfrac{1}{{\sqrt 2 }}$

Answer 1

Hint: To answer the distance from the given point to the given line along the another given line can be calculated by just finding firstly the intersection point of both the lines then finding the distance between the intersection point and the given point using the distance formula between two points.

Complete step-by-step answer:
Given two lines
Suppose
 \[2x - 3y + 9 = 0\;\] ……..(i),
 \[x - y + 1 = 0\] ………(ii)
we will first find the intersection point of the lines
so,
Multiply equation (ii) by 2 and subtract it from equation (1)
 \[2x - 3y + 9 = 0\;\]
 \[2x - 2y + 2 = 0\]
______________
 \[\;{\text{ }}\;{\text{ }}\;\;\; - y + 7 = 0\]
_____________
 \[y = 7,x = 6\]
therefore point of intersection of both lines is \[\left( {6,7} \right)\]
now if we will find the distance between these two points that will be equal to the distance of the point \[\left( {2,3} \right)\;\] from the line \[2x - 3y + 9 = 0\;\] along the line \[x - y + 1 = 0\]
so by using the distance formula
we get the distance between \[\left( {2,3} \right)\;\] and \[\left( {6,7} \right)\]
 \[ = \sqrt {{{\left( {6 - 2} \right)}^2} + {{\left( {7 - 3} \right)}^2}} \]
$ 4\sqrt2 $units
∴ Distance of the point \[\left( {2,3} \right)\;\] from the line \[2x - 3y + 9 = 0\;\] measured along line \[x - y + 1 = 0\] is $ 4\sqrt2 $​ units.
So, the correct answer is “$ 4\sqrt2 $ units”.

Note: In this problem it is not said to find the perpendicular distance from the given point to the given line here it is said that the distance from the given point from the given line along another line. So students are not confused and find the perpendicular distance.