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Class 10 Class 12

Let P and Q be the points of trisection of the line segment joining the points A(2, -2) and B(-7, 4) such that P is nearer to A. Find the coordinates of P and Q.


P and Q are the points of trisection of AB,
therefore, AP=PQ=QB
Thus, P divides AB internally in the ratio 1:2 and Q divide AB internally in the ratio 2:1.

straight P identical to space open parentheses fraction numerator 1 left parenthesis negative 7 right parenthesis plus 2 left parenthesis 2 right parenthesis over denominator 1 plus 2 end fraction comma fraction numerator 1 left parenthesis 4 right parenthesis plus 2 left parenthesis negative 2 right parenthesis over denominator 1 plus 2 end fraction close parentheses identical to space open parentheses fraction numerator negative 7 plus 4 over denominator 3 end fraction comma fraction numerator 4 minus 4 over denominator 3 end fraction close parentheses
identical to open parentheses fraction numerator negative 3 over denominator 3 end fraction comma 0 close parentheses space identical to open parentheses negative 1 comma 0 close parentheses
straight Q identical to space open parentheses fraction numerator 2 left parenthesis negative 7 right parenthesis plus 1 left parenthesis 2 right parenthesis over denominator 2 plus 1 end fraction comma fraction numerator 2 left parenthesis 4 right parenthesis plus 1 left parenthesis negative 2 right parenthesis over denominator 2 plus 1 end fraction close parentheses identical to open parentheses fraction numerator negative 14 plus 2 over denominator 3 end fraction comma fraction numerator 8 minus 2 over denominator 3 end fraction close parentheses
identical to open parentheses fraction numerator negative 12 over denominator 3 end fraction comma 6 over 3 close parentheses space identical to left parenthesis negative 4 comma 2 right parenthesis

8136 Views

Show that the points (-2, 3), (8, 3) and (6, 7) are the vertices of a right
triangle


The given points are  A( -2, 3 )  B( 8, 3 )  and C( 6, 7 ).using distance firmula, we have:AB2 =( 8 - ( -2 ))2 +( 3 - 3 )2AB2 = ( 8+ 2 )2 + 0AB2 = 102AB2 = 100BC2 = (6-8 )2 + (7-3)2 BC2 = ( - 2 )2 + 42BC2 = 4+16 BC2 = 20CA2 = ( -2 -6 )2 + ( 3-7 )2 CA2 = (-8)2 +(-4 )2CA2 = 64 + 16CA2 =80It can be observe that:BC2 + CA2 = 20 + 80 =100 = AB2So, by the converse of pythagoras Theorem,ABC is a right angled triangle  with right angled at C.


If a≠b≠0, prove that the points (a, a2), (b, b2) (0, 0) will not be collinear.


Let A(a, a2), B(b, b2) and C(0, 0) be the coordinates of the given points.
We know that the area of a triangle having vertices (x1, y1), (x2, y2) and (x3, y3) is ∣∣12[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]∣∣ square units.
So,
Area of ∆ABC\
open vertical bar 1 half open square brackets straight a left parenthesis straight b squared minus 0 right parenthesis plus straight b left parenthesis 0 minus straight a squared right parenthesis plus 0 left parenthesis straight a squared minus straight b squared right parenthesis close square brackets close vertical bar
space equals space open vertical bar 1 half left parenthesis ab squared minus straight a squared straight b right parenthesis close vertical bar
space equals 1 half open vertical bar ab left parenthesis straight b minus straight a right parenthesis close vertical bar
not equal to 0 space left parenthesis therefore straight a not equal to straight b not equal to 0 right parenthesis
Since the area of the triangle formed by the points (a, a2), (b, b2) and (0, 0) is not zero, so the given points are not collinear.

7792 Views

The diagonal of a rectangular field is 16 metres more than the shorter side. If the longer side is 14 metres more than the shorter side, then find the lengths of the sides of the field.



Let the shorter side of the rectangular field be x m.

Then, diagonal of the rectangular field = ( x+ 16) m

Also, longer side of the rectangular field = (x +14) m

In right ΔABC

left parenthesis AB right parenthesis squared space plus left parenthesis BC right parenthesis squared space equals space left parenthesis AC right parenthesis squared

rightwards double arrow space left parenthesis straight x space plus 14 right parenthesis squared space plus space straight x squared space equals space left parenthesis space straight x space plus space 16 right parenthesis squared

rightwards double arrow space straight x squared space plus 196 space plus 28 straight x space plus space straight x squared space equals space straight x squared space plus 256 space plus space 32 straight x

rightwards double arrow space straight x squared space minus 4 straight x space minus 60 space equals space 0

rightwards double arrow space straight x squared space minus 10 straight x space plus space 6 straight x minus 60 space equals space 0

rightwards double arrow straight x left parenthesis straight x space minus space 10 right parenthesis space plus space 6 space left parenthesis straight x space minus 10 right parenthesis space equals space 0

rightwards double arrow space straight x space plus space 6 space equals space 0 space or space straight x minus 10 space equals 0

rightwards double arrow space straight x space equals negative 6 space or space space straight x space equals space 10

Since length cannot be negative, so x = 10

therefore, the length of the shorter side = 10 m

Length of the diagonal = 10 +16 = 26m

Length of the longer side = 10 +14 = 24

Hence, the length of the sides of the rectangular field is 10 m and 24 m.
3755 Views

Find the values of k so that the area of the triangle with vertices (1,-1), (-4, 2k) and (-k, 5) is 24 sq. units


The vertices of the given ΔABC are A(1 ,-1), B( -4, 2k ) and C (-k, 5) 

Area of ΔABC =

equals 1 half space left square bracket straight x subscript 1 space left parenthesis straight y subscript 2 space minus straight y subscript 3 right parenthesis space plus space straight x subscript 2 space left parenthesis straight y subscript 3 minus straight y subscript 1 right parenthesis space plus space straight x subscript 3 left parenthesis straight y subscript 1 space minus straight y subscript 2 right parenthesis right square bracket

equals 1 half left square bracket 1 space left parenthesis 2 straight k plus space 5 right parenthesis space minus 4 left parenthesis negative 5 space plus 1 right parenthesis space minus space straight k left parenthesis space minus 1 space minus 2 straight k right parenthesis right square bracket

equals 1 half left parenthesis 2 straight k space plus space 5 space plus space 16 space plus straight k space plus 2 straight k squared right parenthesis

equals 1 half left parenthesis 2 straight k squared space plus 3 straight k space plus 21 right parenthesis space saqure space units
It space is space given space that
Area space of space increment ABC space equals space 24 space square space units
therefore 1 half space left parenthesis 2 straight k squared space plus space 3 straight k space plus 21 right parenthesis equals space 48
rightwards double arrow space 2 straight k squared space plus 3 straight k space plus 21 space equals space 48 space

rightwards double arrow space 2 straight k squared space plus 9 straight k space minus 6 straight k space equals space 0

rightwards double arrow space straight k space left parenthesis 2 straight k space plus 9 right parenthesis space minus 3 space left parenthesis 2 straight k space plus 9 right parenthesis space equals space 0

rightwards double arrow left parenthesis straight k minus space 3 right parenthesis space left parenthesis 2 straight k space plus space 9 right parenthesis space equals space 0

rightwards double arrow straight k minus 3 space equals space 0 space or space 2 straight k space plus 9 space equals space 0

rightwards double arrow space straight k minus space 3 space equals space 0 space or space 2 straight k space plus 9 space equals space 0

rightwards double arrow straight k space equals 3 space or space straight k space equals space minus 9 over 2
Thus space value space of space straight k space is space 3 space or space fraction numerator negative 9 over denominator 2 end fraction

1054 Views

If the area of triangle ABC formed by A(x, y), B(1, 2) and C(2, 1) is 6 square units, then prove that  x + y = 15.


The given vertises are A(x,y), B(1,2) C(2,1).It is know that the area of a tringle whose vertises are (x1,y1). (x2, y2 ), and (x3,  y3)  is given by12 x1 (y2-y3 ) + x2 (y3-y1) + x3(y1-y2)  Area of ABC  = 12 x(2-1) + 1(1-y) +2(y-2)  =12 x+1-y+2y-4 =12 x+y-3The area of ABC is given as 6 sq.units.12x+y-3 = 6x+y-3 = 12x+y=15