A line with slope $4$ intersects a line with slope $7$ at the point $(10,28)$. What is the distance between the $x$-intercepts of these two lines?
This question was asked in a Math Competition in New Jersey, USA. I found $x$-intercepts to be $(6,0)$ and $(3,0)$. The distance is $16$. I do not understand how the distance between the $x$ intercepts is $3$. Please explain. I am only in Grade 6 (Year 6).
Best Answer
A line with slope $4$ is of the form $y=4x+a$ for some constant $a$. A line with slope $7$ is of the form $y=7x+b$ for some constant $b$.
Both lines pass through the point $(x,y) = (10,28)$ and so $28=4\times 10+a$ meaning that $a=-12$. Moreover, $28=7\times 10 + b$ meaning that $b=-42$. The equations of our lines are then $y=4x-12$ and $y=7x-42$.
The $x$-intercept of $y=4x-12$ is given by $0=4x-12$, i.e. $x=3$. The $x$-intercept of $y=7x-42$ is given by $0=7x-42$, i.e. $x=6$. The distance between the points $(3,0)$ and $(6,0)$ is $3$.