Question

How do you find the intercepts for $x + 2y = 10$?

Answer 1

First of all this is a very simple and a very easy problem. The general equation of a straight line is $y = mx + c$, where $m$ is the gradient and $y = c$ is the value where the line cuts the y-axis. The number $c$ is called the intercept on the y-axis. Based on this provided information we try to find the value of the slope and the intercept of the given straight line.

Complete step by step solution:
Consider the given linear equation, as given below:
$\Rightarrow x + 2y = 10$
Convert it into standard line equation form:
$\Rightarrow y = - \dfrac{1}{2}x + 5$
Here this linear equation is in the standard form of the general equation of a straight line.
The general equation of a straight line is given by:
$\Rightarrow y = mx + c$
Here $m$ is the slope of the straight line.
Whereas $c$ is the y-intercept, as it intersects the y-axis at $c$.
So we have to find the slope and the intercept of the given straight line $y = - \dfrac{1}{2}x + 5$.
The slope of the straight line $y = - \dfrac{1}{2}x + 5$, on comparing with the straight line $y = mx + c$,
Here the slope is $m$, and here on comparing the coefficients of $x$,
$\Rightarrow m = - \dfrac{1}{2}$
So the slope of the given straight line $y = - \dfrac{1}{2}x + 5$ is $- \dfrac{1}{2}$.
Now finding the intercept of the line $y = - \dfrac{1}{2}x + 5$, on comparing with the straight line $y = mx + c$, Here the intercept is $c$, and here on comparing the constants of the straight lines,
$\Rightarrow c = 5$
So the intercept of the given straight line $y = - \dfrac{1}{2}x + 5$ is 5.
The slope and intercept of $y = - \dfrac{1}{2}x + 5$ is $- \dfrac{1}{2}$ and $5$ respectively.

Note: Please note that while solving such kind of problems, we should understand that if the y-intercept value is zero, then the straight line is passing through the origin, which is in the equation of $y = mx + c$, if $c = 0$, then the equation becomes $y = mx$, and this line passes through the origin, whether the slope is positive or negative.