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 $$4x+3y=10$$.


ANSWER

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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