Mathematics

Find the sum of ordinates of  all the points on the line $x+y=4$ that lie at a unit distance from the line $4x+3y=10$.

12

SOLUTION
Let the coordinates of point be $(x_1,y_1)$.

Then using the formula of length of perpendicular$[d=|\dfrac{ax_1+by_1+c}{\sqrt(a^2+b^2}|]$:

$1=|\dfrac{4x_1+3y_1-10}{\sqrt{(4^2+3^2)}}|$

$\implies 5+10=4x_1+3y_1$ .......[Eq. 1] or

$\implies -5+10=4x_1+3y_1$..  .... [Eq. 2]

Also $(x_1,y_1)$ lies on line $x+y=4$, hence
$x_1+y_1=4$......[Eq. 3]

So coordinates of required point is intersection of equation [ 1 & 3] and equation [2 & 3].

Hence intersection of eq. (1) and (3) is $A(3,1)$
And intersection of eq. (2) and (3) is $B(-7,11)$.

Hence sum of ordinates of $A$ and $B$ is $[11+1]=12$.

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