BlogIIT-JEEImportant Straight Line Formulas For JEE

Important Straight Line Formulas For JEE

Formulas for the straight line

1. Distance formula:

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    d = √[(x2-x1)2+(y2-y1)2]

    2. Section Formula:

    x = (mx2+nx1)/(m+n)

    y = (my2+ny1)/(m+n)

    3. Centroid:

    G = [(x1+x2+x3)/3, (y1+y2+y3)/3]



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    4. Incentre:

    I = {(ax1+bx2+cx3)/(a+b+c), (ay1+by2+cy3) / (a+b+c)}

    5. Excentre:

    I1 = {(-ax1+bx2+cx3)/(-a+b+c), (-ay1+by2+cy3)/(-a+b+c)}

    6. Slope formula:

    (i) Line joining two points (x1, y1) and (x2, y2), m = (y1 – y2) / (x1 – x2)

    (ii) Slope of a line ax+by+c = 0 is -coefficient of x/coefficient of y = -a/b


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    7. Equation of a straight line in various forms:

    (i) Point Slope form: y-y1 = m(x – x1)

    (ii) Slope intercept form: y = mx + c

    (iii) Two point form: y-y1 = {(y2 – y1) / (x2 – x1)} × (x-x1)

    (iv) Intercept form: (x/a) + (y/b) = 1

    (v) Perpendicular / Normal form: x cos α +y sin α = p

    (vi) Parametric form: x = x1+ r cos θ , y = y1 + r sin θ

    (vii) Symmetric form: (x – x1)/cos θ = (y – y1) / sin θ = r

    (viii) General form: ax + by + c = 0

    x intercept = -c/a

    y intercept = -c/b

    8. Parallel lines:

    Two lines ax+by+c = 0 and a’x+b’y+c’ = 0 are parallel if a/a’ = b/b’ ≠ c/c’.

    Thus any line parallel to ax+by+c = 0 is of the type ax+by+k = 0, where k is a parameter.

    9. Perpendicular lines:

    Two lines ax+by+c = 0 and a’x+b’y+c’ = 0 are perpendicular if aa’+bb’ = 0

    10. Position of the points (x1, y1) and (x2, y2) relative to the line ax+by+c = 0:

    In general, two points (x1, y1) and (x2, y2) will lie on the same side or opposite side of ax+by+c = 0 according to ax1+by1+c and ax2+by2+c are of the same or opposite sign respectively.

    11. Length of the perpendicular from a point on a line :

    The length of the perpendicular from a point (x1, y1) to a line ax + by + c = 0 is

    12. Reflection of a point about a line:

    (i) Foot of the perpendicular from a point on the line is (x-x1)/a = (y-y1)/b = -(ax1+by1+c)/(a2+b2)

    (ii)Image of (x1, y1) in the line ax+by+c = 0 is (x-x1)/a = (y-y1)/b = -2 (ax1+by1+c)/(a2+b2)


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    FAQs:

    Is straight line significant for JEE?

    Yes, it is still in JEE prospectus. It is one of the significant subjects of Coordinate Geometry, so don't miss it.

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