We will understand x intercept in this page.
To understand x intercept, consider the following figure: In the above figure, there is a straight line L which passes through the point A(a, 0) on the x axis. The x coordinate in the point A through which the straight line L passes is called the x intercept of the line L. Unless a straight line L passes through a point lying on the x axis, the x coordinate in the point will not turn into x intercept.
Therefore, x intercept exists with respect to a straight line. Since the x intercept is x coordinate, therefore it can be both positive and negative based on the x coordinate. If the straight line cuts the x axis at the point which lies to the right of the origin O(0, 0), then the x intercept of the line L will be positive.
For example, if a straight line L passes through the point A(3, 0), then the x intercept 3 of the Line L is positive. Look at the figure above for this. And again, if some other straight line Lā passes through the point A (-3, 0), then the x intercept of the line L will be negative. It is because the point A (- 3, 0) lies to the left of the origin O (0, 0). Now, we will discuss some methods for finding x intercept of straight line L based on the information given.
What is the x intercept of the straight line L having a slope 2 and passing through the point P (0, 4)?
We know the x intercept is x coordinate in the point on the x axis trough which a straight line will pass. So, let the point be written as A (a, 0). Now since, the line L passes through the point P (0, 4) also besides A (a, 0), therefore, the slope formula can be applied to find the x intercept a of the line L. From slope formula, we have (4 ā 0)/(0 ā a) = 2, i.e. 4 = -2a, so a = -2
Alternative method. x intercept can also be found using the slope-intercept form of equation of a straight line L we know, the slope-intercept form of equation of a straight line L having slope m and y intercept b is y = mx + b since, the line L passes through the point P (0, 4), therefore, 4 is y intercept of the line L, i,e,. b = 4 in y = mx +b so, the equation becomes y = mx + 4.
And again, substitute the coordinates of the point A (a, 0) as the line L passes through this point. So, we get 0 = 2 Ć a + 4, i.e. 2a = – 4, i.e. a = -2