{"id":6297,"date":"2021-10-12T16:47:38","date_gmt":"2021-10-12T11:17:38","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6297"},"modified":"2021-11-30T16:03:03","modified_gmt":"2021-11-30T10:33:03","slug":"intercept-form-of-a-line-equation","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/intercept-form-of-a-line-equation\/","title":{"rendered":"Intercept Form of a Line Equation"},"content":{"rendered":"
Here you will learn intercept form of a line equation i.e. intercept cut by line on x-axis and y-axis with examples.<\/p>\n
Let’s begin –\u00a0<\/p>\n
\nThe equation of a line which cuts off intercepts a and b respectively from the x and y-axes is<\/p>\n
\\(x\\over a\\) + \\(y\\over b\\) = 1.<\/p>\n<\/blockquote>\n
Proof<\/strong> : Let AB be the line which cuts off intercepts OA = a and OB = b on the x and y axes respectively.\u00a0<\/p>\n
Let P(x,y) be any point on the line. Draw PL \\(\\perp\\) OX.<\/p>\n
Then, OL = x and PL = y.<\/p>\n
Clearly,\u00a0<\/p>\n
Area of OAB = Area of triangle OPA + Area of triangle OPB<\/p>\n
\\(1\\over 2\\) OA.OB = \\(1\\over 2\\) OA.PL + \\(1\\over 2\\) OB.PM<\/p>\n
\\(1\\over 2\\)ab = \\(1\\over 2\\)ay + \\(1\\over 2\\)bx<\/p>\n
\\(\\implies\\) ab = ay + ax<\/p>\n
\\(\\implies\\) \\(x\\over a\\) + \\(y\\over b\\) = 1.<\/p>\n
This is the required equation of the line in the intercept form.<\/p>\n
Example<\/span><\/strong> : Find the equation of the line which cut off an intercept 4 on the positive direction of x-axis and an intercept 3 on the negative direction of y-axis.<\/p>\n
Solution<\/span><\/strong> : Here a = 4, b = -3.<\/p>\n
So, the equation of the line is<\/p>\n
\\(x\\over a\\) + \\(y\\over b\\) = 1 or, \\(x\\over 4\\) + \\(y\\over -3\\) = 1. or 3x – 4y = 12.<\/p>\n
Example<\/span><\/strong> : Find the equation of the straight line which passes through the point (4, -2) and whose intercept on y-axis is twice that on X-axis.<\/p>\n
Solution<\/span><\/strong> : let the equation of line be<\/p>\n
\\(x\\over a\\) + \\(y\\over b\\) = 1\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0………….(i)<\/p>\n
It is given that its y-intercept is twice the x-intercept<\/p>\n
\\(\\therefore\\)\u00a0 b = 2a<\/p>\n
Putting b = 2a in (i), we get<\/p>\n
2x + y = 2a\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0………………..(ii)<\/p>\n
It passes through point (4, -2).<\/p>\n
\\(\\therefore\\)\u00a0 8 – 2 = 2a \\(\\implies\\) a = 3.<\/p>\n
Substituting a = 3 in (ii), we get<\/p>\n
2x + y = 6 as the equation of the required line.<\/p>\n\n\n