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November 22, 2024Intercept: An intercept is a point on the \(Y\)-axis through which the slope of a line passes in mathematics. It is the \(Y\)-coordinate of a location on the \(Y\)-axis where a straight line or a curve meets. When we write the equation for a line, \(y=mx+c\), where \(m\) is the slope and \(c\) is the \(y\)-intercept, we get this.
The \(x\)-intercept and \(y\)-intercept are the two types of intercepts. The \(x\)-intercept is the point where the line crosses the \(X\)-axis, while the \(y\)-intercept is the point where the line crosses the \(Y\)-axis. In this article, we will go through the concept of intercept in depth.
The graphical representation of \(x\)- and \(y\)-intercepts is straightforward. Where the graph crosses the \(X\)-axis are the \(x\)-intercepts, and where the graph crosses the \(Y\)-axis are the \(y\)-intercepts. When we try to deal with intercepts algebraically, we run into difficulties.
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Consider the axes once again to explain the algebraic component. When you first learned about the Cartesian plane in primary school, you were shown how to draw a normal number line (the \(X\)-axis) and create a perpendicular number line (the \(Y\)-axis) through the zero points on the first number line. If you look closely, you’ll notice that the \(Y\)-axis is also the line \(“x=0.”\) The \(X\)-axis is also the line \(“y=0″\) in this case.
Then an \(x\)-intercept is a point on the graph where \(y\) is zero, and a \(y\)-intercept is a point where \(x\) is zero, algebraically speaking. An \(x\)-intercept is a position in an equation where the \(y\)-value is zero, while a \(y\)-intercept is a point where the \(x\)-value is zero.
An intercept is a point on a graph where a line or curve crosses across the axis. The \(x\)-intercept, or horizontal intercept, is a point that crosses the \(X\)-axis. And that point is known as the \(y\)-intercept or vertical intercept when it crosses the \(Y\)-axis.
Let’s get familiar with the terminology required to calculate the horizontal and vertical intercepts.
The first step is to figure out what the line’s equation is. A straight line’s equation is written as \(y=mx+b.\) The slope of the line is denoted by \(m\), while the \(y\)-intercept is denoted by \(b\).
We should now understand what the slope of a line means. Its inclination or steepness determines the slope of a line. It’s the proportion of \(y\) value change to \(x\) value change.
It’s known as “rising over run.”
\({\rm{Slope = }}\frac{{{\rm{ change\, in\, y\, value }}}}{{{\rm{ change\, in\, x \,value }}}}{\rm{ = }}\frac{{{\rm{ Rise }}}}{{{\rm{Run}}}}\)
Slope \(=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
By choosing any two points in the graph above, we can find the increase in the graph, which is \(4\) in this case. The slope here is \(\frac{4}{3}\) since the run is \(3\).
The slopes are classified as follows:
Suppose we have two points \(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right)\) then the formula to find the equation of a line is given by
\(\left(y-y_{1}\right)=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\left(x-x_{1}\right)\)
The \(y\) value of the place where the line crosses the \(Y\)-axis is called the \(y\)-intercept. By looking at the graph and seeing where the \(Y\)-axis is severed, we may identify the \(y\)-intercept. This point’s \(x\)-coordinate will always be \(0\). It will be formatted as follows: \((0,y)\). Because it is located on the vertical axis, it is also known as the vertical intercept.
Let \(x=0\) before solving for \(y\) to get the \(y\)-intercept of an equation.
The \(x\)-intercept is the point on a graph of a function or equation where it intersects with the \(X\)-axis. This may be thought of as a point with a value of \(0\). Because it is located on the horizontal axis, it is also known as the horizontal intercept.
Let \(y=0\) before solving for \(x\) to get the \(x\)-intercept of an equation. The \(x\)-intercept coordinates will be \((x,0)\).
Example: Find \(x\) and \(y\) intercepts for the line \(6x+3y=18.\)
To find the \(x\)-intercept put \(y=0\)
\(6x+3(0)=18\)
\(\Rightarrow 6 x=18\)
\(\Rightarrow x=3\)
So, the \(x\)-intercept is \((3,0)\).
To find the \(y\)-intercept put \(x=0\)
\(6(0)+3y=18\)
\(\Rightarrow 3 y=18\)
\(\Rightarrow y=6\)
So, the \(y\)-intercept is \(0,6\).
The \(x\) and \(y\) intercepts can now be determined by looking at the graphs. Try to locate the \(x\) and \(y\) intercepts in the graph below.
The \(x\)-intercept is the point where the line crosses the \(x\)-axis. It’s \(-4\). Therefore the \(x\)-intercept is \((-4,0)\)
The \(y\)-intercept is now the point at which it crosses the \(Y\)-axis.
Because it crosses the \(Y\)-axis at \(6\), the \(y\)-intercept is \((0,6)\)
The intercepts can also be used to graph lines. Let’s look at an example of this idea.
Example: Find the \(x\) and \(y\) intercepts to make the graph of \(5x+3y=15\).
To find the \(x\)-intercept put \(y=0\)
\(5x+3(0)=15\)
\(\Rightarrow 5 x=15\)
\(\Rightarrow x=3\)
So, the \(x\)-intercept is \((3,0)\).
To find the \(y\)-intercept put \(x=0\)
\(5(0)+3y=15\)
\(\Rightarrow 3 y=15\)
\(\Rightarrow y=5\)
So, the \(y\)-intercept is \((0,5)\).
Draw a line joining the \(x\) and \(y\) intercepts.
Keep in mind that the \(x\) and \(y\) intercepts are ordered pairs, not integers. If the value of \(x\) intercept is \(2\), the expression is written as \((2,0)\). Furthermore, not every graph contains both horizontal and vertical intercepts. Consider the following scenario:
There is no \(x\) intercept on the horizontal line shown above. As far as we can see, it just has a \(y\)-intercept \((0,-2)\).
Let’s look at a situation when there is no \(y\)-intercept.
The above vertical line has an \(x\) intercept of \((3,0)\) but no \(y\)-intercept.
Now consider the scenario where the \(x\) and \(y\)-intercepts are the same, i.e., origin.
After determining the \(x\) and \(y\)-intercepts, graph: \(y=-2x\).
\(x=0\) is used to find the \(x\)-intercept.
\(y=-2(0)\)
\(y=0\)
So, the \(y\)-intercept is \((0,0)\)
Put \(y=0\) to determine the \(x\)-intercept.
\(0=-2x\)
\(\Rightarrow x=0\)
Thus, the \(x\)-intercept is \((0,0)\)
As a result, the \(x\) and \(y\) intercepts are the same locations, the origin. To graph the line, we will need one more point. Determine the value for \(5\) using any \(x\) value.
When we set \(x\) to \(1\), we obtain \(y=-2\).
When we set \(x\) to \(-1\), we obtain \(y=2\).
Now graph the line \((0,0),(-1,2),(1,-2)\) using these ordered pairs.
Q.1. Find x and y intercepts for the line 3x-y=9.
Ans: To find the \(x\)-intercept put \(y=0\)
\(3x-(0)=9\)
\(\Rightarrow 3 x=9\)
\(\Rightarrow x=3\)
So, the \(x\)-intercept is \((3,0)\).
To find the \(y\)-intercept put \(x=0\)
\(3(0)+y=9\)
\(\Rightarrow y=9\)
\(\Rightarrow y=9\)
So, the \(y\)-intercept is \((0,9.)\)
Q.2. Find the y intercept for the line x+3y=12.
Ans: Given, \(x+3y=12\)
To find the \(y\)-intercept put \(x=0\)
\((0)+3y=12\)
\(\Rightarrow 3 y=12\)
\(\Rightarrow y=4\)
So, the \(y\)-intercept is \(0,4.\)
Q.3. Write the equation if x and y intercepts for the line are 3,0 and 0,5 respectively.
Ans: Given, the \(x\) intercept is \((3,0)\) and the \(y\)-intercept is \(0,5\)
So by using the formula
\(m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
We will find the slope of the line
\(m=\frac{5-0}{0-3}=-\frac{5}{3}\)
Putting the value of slope in the equation \(y=mx+c\), we get
\(y=-\frac{5}{3} x+5\)
\(3 y=-5 x+15\)
Or \(5 x+3 y=15\)
Hence, the required equation is \(5x+3y=15\).
Q.4. Find x intercept for the line 2x+y=1.
Ans: To find the \(x\)-intercept put \(y=0\)
\(2x+(0)=1\)
\(\Rightarrow 2 x=1\)
\(\Rightarrow x=\frac{1}{2}\)
So, the \(x\)-intercept is \(\left(\frac{1}{2}, 0\right)\)
Q.5. Find x and y intercepts for the line 16x+5y=20.
Ans: To find the \(x\)-intercept put \(y=0\)
\(16x+5(0)=20\)
\(\Rightarrow 16 x=20\)
\(\Rightarrow x=\frac{20}{16}=\frac{5}{4}\)
So, the \(x\)-intercept is \(\left(\frac{5}{4}, 0\right)\)
To find the \(y\)-intercept put \(x=0\)
\(16(0)+5y=20\)
\(\Rightarrow 5 y=20\)
\(\Rightarrow y=4\)
So, the \(y\)-intercept is \(0,4\).
We went through the basics of intercepts in this article, followed by a discussion of intercept definitions. The slope-intercept form of a line and the two-point form of a line was then discussed. After that, we used a graph to demonstrate the concepts of \(x\) and \(y\)-intercepts and certain instances where there are no \(x\) and \(y\)-intercepts. Finally, we looked at several solved cases to obtain a better understanding of the subject.
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Q.1. What is the intercept in maths?
Ans: The point at which a line or curve crosses the graph’s axis is the intercept. The term \(x\)-intercept refers to the point that crosses the \(X\)-axis. The term \(y\)-intercept refers to the point that crosses the \(Y\)-axis.
Q.2. What is the equation of the line with respect to x and y-intercepts?
Ans: When we write the equation for a line, \(y=mx+c\), where m is the slope and \(c\) is the \(y\)-intercept, we get this. The \(x\)-intercept and \(y\)-intercept are the two types of intercepts. The \(x\)-intercept is the point where the line crosses the \(X\)-axis, while the \(y\)-intercept is the point where the line crosses the \(Y\)-axis.
Q.3. How do you find the intercept?
Ans: We put \(y=0\) and solve for \(x\) to find the \(x\)-intercept. Similarly, we put \(x=0\) and solve for \(y\) to find the \(y\)-intercept.
Q.4. What is the y-intercept when given two points?
Ans: Suppose we have two given points \(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right)\) then first, we find the slope by using the formula
\(m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Then put the value of the slope in the slope-intercept form of the equation \(y=mx+c\)
Substitute either of the two points in the slope-intercept form. Then solve for \(c\), which is the \(y\)-intercept of the equation.
Finally, substitute the value of \(c\) into the equation.
Q.5. What is the slope formula?
Ans: The formula to find the slope of a line is
Slope \(=\frac{\text { change in } y \text { value }}{\text { change in } x \text { value }}\)
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