{"id":3888,"date":"2021-08-09T21:01:28","date_gmt":"2021-08-09T21:01:28","guid":{"rendered":"https:\/\/mathemerize.com\/?p=3888"},"modified":"2021-11-30T16:37:53","modified_gmt":"2021-11-30T11:07:53","slug":"equation-of-straight-lines","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/equation-of-straight-lines\/","title":{"rendered":"Equation of Straight Lines in all Forms"},"content":{"rendered":"
Here, you will learn equation of straight lines in all forms i.e. slope form, intercept form, normal form and parametric form etc.<\/p>\n
Let’s begin – <\/p>\n
A relation between x and y which is satisfied by co-ordinates of every point lying on a line is called equation of the straight lines. Here, remember that every one degree equation in variable x and y always represent a straight line i.e. ax + by + c = 0 ; a & b \\(\\ne\\) 0 simultaneously.<\/p>\n
\n(a) Equation of a line parallel to x-axis at a distance ‘a’ is y = a or y = -a.<\/p>\n
(b) Equation of x-axis is y = 0<\/p>\n
(c) Equation of a line parallel to y-axis at a distance ‘b’ is x = b or x = -b.<\/p>\n
(d) Equation of y-axis is x = 0.<\/p>\n<\/blockquote>\n
Different Equation of Straight Lines <\/h2>\n
(a) Slope Intercept form :<\/strong><\/h4>\n
Let m be the slope of a line and c its intercept on y-axis. Then the equation of this straight line is written as<\/p>\n
\ny = mx + c.<\/p>\n<\/blockquote>\n
(b) Point Slope form :<\/strong><\/h4>\n
Let m be the slope of a line and it passes through a point (\\(x_1,y_1\\)), then its equation is written as :<\/p>\n
\ny – \\(y_1\\) = m(x – \\(x_1\\)).<\/p>\n<\/blockquote>\n
(c) Two Point form :<\/strong><\/h4>\n
Equation of a line passing through two points (\\(x_1,y_1\\)) and (\\(x_2,y_2\\)) is written as <\/p>\n
\ny – \\(y_1\\) = \\(y_2-y_1\\over {x_2-x_1}\\)(x – \\(x_1\\)).<\/p>\n<\/blockquote>\n
(d) Intercept form :<\/strong><\/h4>\n
If a and b are the intercepts made by a line on the axes of x and y, its equation is written as :<\/p>\n
\n\\(x\\over a\\) + \\(y\\over b\\) = 1<\/p>\n
Length of intercept of line between the coordinate axes = \\(\\sqrt{a^2+b^2}\\)<\/p>\n<\/blockquote>\n
(e) Normal form :<\/strong><\/h4>\n
If p is the length of perpendicular on a line from the origin, and \\(\\alpha\\) the angle which this perpendicular makes with positive x-axis, then the equation of this line is written as<\/p>\n
\nxcos\\(\\alpha\\) + ysin\\(\\alpha\\) = p (p is always positive) where 0 \\(\\le\\) \\(\\alpha\\) < 2\\(\\pi\\).<\/p>\n<\/blockquote>\n
(f) Parametric form :<\/strong><\/h4>\n
To find the equation of a straight line which passes through a given point A(h,k) and makes a given angle \\(\\theta\\) with the positive direction of the axis. P(x,y) is any point on the line.<\/p>\n
\nLet AP = r, then x – h = rcos\\(\\theta\\), y – k = rsin\\(\\theta\\) & \\(x – h\\over {cos\\theta}\\) = \\(y – k\\over {sin\\theta}\\) = r is the equation of straight line.<\/p>\n<\/blockquote>\n
(g) General form :<\/strong><\/h4>\n
We know that a first degree equation in x and y, ax + by + c = 0 always represent a straight line. This form is known as general form of straight line.<\/p>\n
\n(i) Slope of this line = -\\(a\\over b\\)<\/p>\n
(ii) Intercept by this line on x-axis = -\\(c\\over a\\) and Intercept by this line on y-axis = -\\(c\\over b\\)<\/p>\n
(iii) To change the general form of a line to normal form, first take c to right hand side and make it positive, then divide the whole equation by \\(\\sqrt{a^2+b^2}\\).<\/p>\n<\/blockquote>\n\n\n
Example : <\/span> Equation of a line which passes through point A(2,3) and makes an angle of 45 with x-axis. If this line meet the line x + y + 1 = 0 at a point P then distance AP is-<\/p>\n
Solution : <\/span>Here \\(x_1\\) = 2, \\(y_1\\) = 3 and \\(\\theta\\) = 45
\n Hence \\(x-2\\over cos45\\) = \\(y-3\\over sin45\\) = r
\n from first two parts \\(\\implies\\) x – 2 = y – 3 \\(\\implies\\) x – y + 1 = 0
\n Co-ordinate of point P on this line is (2+\\(r\\over \\sqrt{2}\\), 3+\\(r\\over \\sqrt{2}\\)).
\n If this point is on line x + y + 1 = 0 then
\n (2+\\(r\\over \\sqrt{2}\\)) + (3+\\(r\\over \\sqrt{2}\\)) + 1 = 0 \\(\\implies\\) r = -\\(3\\sqrt{2}\\) ; |r| = \\(3\\sqrt{2}\\)
<\/p>\n\n\n\n