Question

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

Answer 1

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

Then using the formula of length of perpendicular$[d=|\frac{a{x}_1+b{y}_1+c}{\sqrt{(}{a}^2+b^2}|]$:

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

$⟹5+10=4x_1+3y_1$ .......[Eq. 1] or

$⟹-5+10=4x_1+3y_1$.. .... [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 $[11+1]=12$.