 Hello and welcome to the session. In this session, we will discuss about the intersets and slope intercept form of a line. First of all, let's see what are intersets. Consider the straight line L axis. There is this axis at the point A. It makes the y-axis at the point B. Then we have x-intercept, the distance of the point where the straight line A meets the x-axis. That is, it meets the x-axis at the point A. The distance of the point A from the origin, which is OA, is the x-intercept, which is the intercept made by the line on the x-axis. Now let's see what is the y-intercept. The y-intercept with the distance of the point B, which is the point where the line A meets the y-axis. So the distance of this point B from the origin, which is OB, is the y-intercept. That is the intercept made by the line L on the y-axis. Now next we discuss gradient, the slope or gradient of a straight line. The part of the line above the x-axis makes with the positive direction the axis. Suppose in this figure we have the straight line L and the straight line L makes angle with the positive direction of the x-axis or the gradient line L is equal to tan theta and the slope is denoted by the letter M. So for the straight line L, M, which is the slope is equal to tangent of the angle, which the part of the line above the x-axis makes with the positive direction of x-axis, that is, angle theta. Next we discuss the slope-intercept form. So there this figure in which we have a straight line L till the x-axis, so at a point A, y-axis at a point B and this straight line L makes an angle theta with the x-axis part of this line which means O B with B that is we have O B is equal to C and we suppose the slope of the line L given by M is equal to tan theta to the point B which coordinates on the line L we draw perpendicular to the x-axis that is O A perpendicular to O X next we also draw perpendicular to P M. So this is B M perpendicular to P M and this figure A P is equal to theta now as A L is tan L to B M so this angle would also be equal to theta that is angle B P would also be equal to theta now let's consider the triangle P B M and theta is equal to that is the perpendicular upon the base that is B M. You can see that the point P has coordinates x is equal to y, the distance of the point P from the y-axis would be x, B M would also be y minus C which means B upon X that M is equal to y minus C upon X is equal to y minus C, y is equal to M and the y-intercept is the slope-intercept form of a line where M which is the coefficient of x y-intercept means we have value of M as minus 2 and value of C as 7, the slope-intercept y equal to M plus the y-intercept that is C which is 7. When the values of M and C we can easily write the equation of the line in the slope-intercept form. In this session we have understood the concept of intercepts and slope-intercept form of a line.