Question

⭐formula is of ➡ Ço-ordinate geometry: Distance formula ⭐




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Answer 1

Distance between two points P(x1,y1) and Q(x2,y2) is given by: d(P, Q) = √ (x2 − x1)2 + (y2 − y1)2 {Distance formula} 2. Distance of a point P(x, y) from the origin is given by d(0,P) = √ x2 + y2. 3.Equation of the x-axis is y = 0 4.

Answer 2

Hi there!
Here's the answer:

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¶¶¶ Coordinate Geometry of
2 Dimension ¶¶¶

¶¶¶¶ THE POINT:

1. Distance formulae :
(a) Distance b/w 2 points (x1,y1) & (x2,y2) is √[ (x2-x1)² + (y2-y1)² ]

(b) Distance of (x,y) from the origin is √(x² + y²)

2.
(a) Co ordinate of the point which divides the line joining (x1,y1) & (x2,y2) internally in ratio m : n is
[(m1x2+ m2x1)/(m1+m2) , (m1y2 + m2y1)/(m1+m2) ]

(b) Co ordinate of the point which divides the line joining (x1,y1) & (x2,y2) externally in ratio m : n is
[(m1x2 - m2x1)/(m1-m2) , (m1y2 - m2y1)/(m1-m2) ]

(c) Coordinate of point which bisects the line joining (x1,y1) & (x2,y2) are :
[(x1 + x2)/2 , (x1 - x2)/2]
i. e., [Mean of x' s , Mean of y' s]

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¶¶¶ TRIANGLE
1.
(a) Coordinate of centroid of triangle whose vertices are (x1, y1) , (x2, y2) & (x3,y3) are:
[ (x1+x2+x3)/3 , (y1+y2+y3)/3 ]

(b) Coordinate of the in centre of the triangle whose vertices are (x1, y1) , (x2, y2) & (x3,y3) are :

[ (ax1+bx2+cx3)/(a+b+c) ; (ay1+by2+cy3)/(a+b+c) ]


2. Area of triangle whose vertices are (x1, y1) , (x2, y2) & (x3,y3) is :
(1/2)×[ (x1y2 - x2y1) + (x2y3 - x3y2) + (x3y1 - x1y3) ]



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¶¶¶ STRAIGHT LINE & INTERCEPTS

• Slope of line (m) is the tangent of the angle which the line makes with positive direction of the x-axis.
If a is the angle made by line with +ve x-axis, then m = tan a

• If a line cuts the x-axis at A & y-axis at B, then OA is called intercept on x-axis & OB is called intercept on y-axis. Both are called intercepts

(a) x = a is the eq. of line //el to y-axis & at a distance "a" from y-axis.
(b) x = 0 is eq. of y- axis
(c) y = a is the eq. of line //el to x-axis & at a distance "a" from x-axis.
(d) x = 0 is eq. of y-axis

(e) y = mx is the eq. of line which passes through the origin and whose slope is "m"

(f) y = mx+c is the equation of line whiae slope is m & c is intercept on y-axis
This is known as Slope form or Tangent form

(g) (x/a) + (y/b) = 1 is the eq.n of a line where "a" is the intercept on x-axis & "b" is the intercept on x-axis
This is Intercept Form

(h) Normal form or perpendicular form
x cos a + y sin a = p is the eq. of a line
where p is the length of perpendicular drawn from (0,0) on the line
a is angle which is perpendicular makes with the +very direction of x-axis

(i) Point slope form :

y - y1 = m(x - x1) is the eq. of line which passes through the point (x1,y1) and whose slope is m

(j) Two points form:

(y-y1) = [ (y2-y1)/(x2-x1) ] × (x-x1)

(k) Distance form:

(x - x1)/cos a = (y-y1)/sin a= r is the eq. of a line which passes through the point (x1,y1) makes an angle a with the x-axis and r is the distance of point (x,y) from (x1,y1)

(l) Any and every eq. of first degree in x and y always represents a straight line
Hence eq. is of form ax+by+c = 0

• Its slope = -(a/b)

(m) Slope of line passing through (x1,y1) & (x2,y2) is
(y2-y1) / (x2-x1)
i.e., [Difference of ordinates ÷ Difference of abscissae]


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¶¶¶¶ CIRCLE

1. x² +y² = a² , is the eq. of circle whose centre is (0,0) and radius "a"

2. (x-h)²+(y-k)² = a² , is the eq. of circle whose centre is (h,k) and radius "a"

3. Diameter form :
(x-x1)(x-x2)+ (y-y1)(y-y2) = 0 is the eq. of circle on the join of (x1,y1) and (x2,y2) as a diameter



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Hope it helps

¶¶¶¶¶¶¶ HAPPY DIWALI ¶¶¶¶¶¶¶