{"id":6293,"date":"2021-10-12T16:42:28","date_gmt":"2021-10-12T11:12:28","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6293"},"modified":"2021-11-30T16:16:31","modified_gmt":"2021-11-30T10:46:31","slug":"what-is-the-point-slope-form-of-a-line-equation","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/what-is-the-point-slope-form-of-a-line-equation\/","title":{"rendered":"What is the Point Slope Form of a Line Equation"},"content":{"rendered":"
Here you will learn what is the point slope form of a line equation with proof and examples.<\/p>\n
Let’s begin – <\/p>\n
\nThe equation of a line which passes through the point \\(P(x_1, y_1)\\) and has the slope ‘m’ is<\/p>\n
\\(y – y_1\\) = m\\((x – x_1)\\)<\/p>\n<\/blockquote>\n
Proof<\/strong> :<\/p>\n
\nLet \\(Q(x_1, y_1)\\) be the point through which the line passes and let P(x, y) be any point on the line.<\/p>\n
Then the slope of the line is \\(y – y_1\\over x – x_1\\)<\/p>\n
but, m is the slope of line <\/p>\n
\\(\\therefore\\) m = \\(y – y_1\\over x – x_1\\) \\(\\implies\\) \\(y – y_1\\) = m(\\(x – x_1\\))<\/p>\n
Hence, \\(y – y_1\\) = m\\(x – x_1\\) is the required equation of the line.<\/p>\n<\/blockquote>\n
Example<\/strong><\/span> : Find the equation of a line passing through (2, -3) and inclined at an angle of 135 with the positive direction of x-axis.<\/p>\n
Solution<\/span><\/strong> : Here, m = slope of the line = tan 135 = tan(90 + 45) = -cot 45 = -1<\/p>\n
\\(x_1\\) = 2, \\(y_1\\) = -3<\/p>\n
So, the equation of the line is<\/p>\n
\\(y – y_1\\) = m(\\(x – x_1\\))<\/p>\n
i.e. y – (-3) = -1(x – 2)<\/p>\n
\\(\\implies\\) y + 3 = -x + 2 \\(\\implies\\) x + y + 1 = 0.<\/p>\n
which is the required equation of line.<\/p>\n
Example<\/strong><\/span> : Determine the equation of line through the point (-4, -3) and parallel to x-axis<\/p>\n
Solution<\/span><\/strong> : Here, m = slope of the line = 0,<\/p>\n
\\(x_1\\) = -4, \\(y_1\\) = -3<\/p>\n
So, the equation of the line is<\/p>\n
\\(y – y_1\\) = m(\\(x – x_1\\))<\/p>\n
i.e. y + 3 = 0(x + 4)<\/p>\n
\\(\\implies\\) y + 3 = 0.<\/p>\n
which is the required equation of line.<\/p>\n\n\n