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Question: Find the distance of the point \[\left( {2,3} \right)\;\] form the line \[2x - 3y + 9 = 0\;\] measur...

Find the distance of the point (2,3)  \left( {2,3} \right)\; form the line 2x3y+9=0  2x - 3y + 9 = 0\; measurement along a line xy+1=0x - y + 1 = 0 .
A.12\dfrac{1}{{\sqrt 2 }}
B.424\sqrt 2
C.2\sqrt 2
D.12\dfrac{1}{{\sqrt 2 }}

Explanation

Solution

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
2x3y+9=0  2x - 3y + 9 = 0\; ……..(i),
xy+1=0x - 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)
2x3y+9=0  2x - 3y + 9 = 0\;
2x2y+2=02x - 2y + 2 = 0


            y+7=0\;{\text{ }}\;{\text{ }}\;\;\; - y + 7 = 0


y=7,x=6y = 7,x = 6
therefore point of intersection of both lines is (6,7)\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 (2,3)  \left( {2,3} \right)\; from the line 2x3y+9=0  2x - 3y + 9 = 0\; along the line xy+1=0x - y + 1 = 0
so by using the distance formula
we get the distance between (2,3)  \left( {2,3} \right)\; and (6,7)\left( {6,7} \right)
=(62)2+(73)2= \sqrt {{{\left( {6 - 2} \right)}^2} + {{\left( {7 - 3} \right)}^2}}
424\sqrt2units
∴ Distance of the point (2,3)  \left( {2,3} \right)\; from the line 2x3y+9=0  2x - 3y + 9 = 0\; measured along line xy+1=0x - y + 1 = 0 is 424\sqrt2​ units.
So, the correct answer is “424\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.