{"id":6291,"date":"2021-10-12T16:40:26","date_gmt":"2021-10-12T11:10:26","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6291"},"modified":"2021-11-30T16:18:11","modified_gmt":"2021-11-30T10:48:11","slug":"what-is-slope-intercept-form-of-a-line-equation","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/what-is-slope-intercept-form-of-a-line-equation\/","title":{"rendered":"What is Slope Intercept Form of a Line Equation"},"content":{"rendered":"
Here you will learn what is slope intercept form of a line with examples.<\/p>\n
Let’s begin – <\/p>\n
\nThe equation of a line with slope m and making an intercept on y-axis is<\/p>\n
y = mx + c<\/p>\n<\/blockquote>\n
Proof<\/strong> :<\/p>\n
Let the given line intersects y-axis at Q and makes an angle \\(\\theta\\) with x-axis. Then, m = \\(tan\\theta\\).<\/p>\n
Let P(x, y) be any point on the line. Draw PL perpendicular to x-axis and QM \\(\\perp\\) PL.<\/p>\n
Clearly, \\(\\angle\\) MPQ = \\(\\theta\\), QM = OL = x<\/p>\n
and, PM = PL – ML = PL – OQ = y – c<\/p>\n
From triangle PMQ, we have<\/p>\n
\\(\\tan\\theta\\) = \\(PM\\over QM\\) = \\(y – c\\over x\\)<\/p>\n
\\(\\implies\\) m = \\(y – c\\over x\\)<\/p>\n
\\(\\implies\\) y = mx + c<\/p>\n
which is the required equation of the line.<\/p>\n
Remark 1<\/strong> : <\/p>\n
\nIf the line passes through the origin, the 0 = m0 + c \\(\\implies\\) c = 0. Therefore, the equation of a line passing through origin is y = mx, where m is the slope of the line.<\/p>\n<\/blockquote>\n
Remark 2<\/strong> : <\/p>\n
\nIf the line is parallel to x-axis, then m = 0, therefore the equation of a line parallel to x-axis is y = c.<\/p>\n<\/blockquote>\n
Example<\/strong><\/span> : Find the equation of a line with slope -1 and cutting off an intercept of 4 units on negative direction of y-axis.<\/p>\n
Solution<\/span><\/strong> : Here m = -1 and c = -4<\/p>\n
Substituting these values in y = mx + c, we obtain that the equation of the line is<\/p>\n
y = – x – 4 or, x + y + 4 = 0<\/p>\n
Example<\/strong><\/span> : Find the equation of a straight line which cut off an intercept of 5 units on negative direction of y-axis and makes an angle of 120 with the positive direction of x-axis.<\/p>\n
Solution<\/span><\/strong> : Here m = tan 120 = tan(90 + 30) = -cot 30 = -\\(\\sqrt{3}\\) and c = -5<\/p>\n
Substituting these values in y = mx + c, we obtain that the equation of the line is<\/p>\n
y = – \\(\\sqrt{3}\\)x – 5<\/p>\n
or, \\(\\sqrt{3}\\)x + y + 5 = 0<\/p>\n\n\n