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The point which divides the joint of (1, 2) and (3,4) externally in the ratio 1 : 1.
x_{1} = (1*3  1*1)/(11)
y_{1} = (1*4  1*2)/(11)
Hence the denominator becomes zero, here we can say that any point cannot be divided in the ratio of 1:1 externally.
Stepbystep explanation: We are given to find the distance of the point (x, y) from the origin.
We know that the coordinates of the origin are (0, 0).
And, the distance between any two points (a, b) and (c, d) is given by the following distance formula:
Therefore, the distance between the point (x, y) and the origin will be
Distance is a metric, a bifunction d, which is always nonnegative, along with being symmetric, satisfies the triangle inequality, and the identity of indiscernibles (i.e., d(x,y)=0⟺x=y)
d(x,y)=0⟺x=y). The “nearest” distance from a point (x1,y1) to the yaxis in 2space is along the line y=y1. (orthogonal to the yaxis). This line interests the yaxis exactly at the point (0,y1). Using the Euclidean distance metric on R^{2}, one obtains:
d((x,y),(0,y))= √(x−0)^{2}+(y−y)^{2}
= x
The line through the points (a , b) and ( a,  b) passes through the point
Slope of line passing through (a,b) and (−a,−b) is given by (b+b)/(a+a) = b/a
So equation of line passing is given by (using slope point form)
y−b = b/a(x−a)
⇒ ay − ab = bx − ab
⇒ ay = bx
Clearly the point (a^{2},ab) lie on the above line
Projection (the foot of perpendicular) from (x , y) on the x – axis is
Slope of a line is not defined if the line is parallel to x axis as all points on the line have same x coordinate.
The distance between the lines 5x – 12y + 65 = 0 and 5x – 12y – 39 = 0
5x − 12y + 65 = 0, 5x − 12y − 39 = 0
Since, these lines are parallel .
a = 5, b = 12, c_{1} = 65 c_{2} = 39
So, d = c1 − c2/[a^{2} + b^{2}]^{1/2}
⇒ d = 104/13
= 8
The equation y−y_{1} = m (x−x_{1}), m ∈ R, represents all lines through the point (x_{1},y_{1}) except the line
The distance between the parallel lines 3x + 4y + 13 = 0 and 3x + 4y – 13 = 0 is
Given parallel lines3x +4y+13=0
perp length from origin P1= 13/√(3)^{2} +(4)^{2}
=13/5
and another line 3x+4y13=0 P2= 13/√(3)^{2}+(4)^{2}
=13/5
Now distance b/w parallel lines
D=P1P2
=13/5+13/5
=26/5 units
The straight lines x + y = 0 , 3x + y – 4 = 0 , x + 3y – 4 = 0 form a triangle which is
Explanation: x+y=0 ..... (i)
3x+y−4=0 ..... (ii)
x+3y−4=0 ....... (iii)
Solving lines (i) and (ii), we get
−x=−3x+4
⟹ x=2, y=−2
∴(i) and (ii) intersect at A=(2,−2)
Solving lines (ii) and (iii), we get
−3x+4 = (4−x)/3
⟹ −9x+12 = 4−x
⟹x=1, y=1
∴(ii) and (iii) intersect at B=(1,1)
Solving lines (i) and (iii), we get
x = (4−x)/3
⟹ −3x = 4−x
⟹ x=−2, y=2
So, AB,BC,AC form a triangle ABC
Now, AB = [(1−2)^{2} + (1+2)^{2}]^{1/2} = (10)^{1/2}
BC = [(−2−1)^{2} + (2−1)2]^{1/2} = (10)^{1/2}
AC = [(−2−2)^{2} + (2+2)^{2}]^{1/2} = (32)^{1/2}
∴ AB=BC
Since, two sides of a triangle are equal then the triangle formed by A,B,C is isosceles triangle.
The equation of the line which passes through the point (1 ,  2) and cuts off equal intercepts from the axis is
If a line is drawn through the origin and parallel to the line x – 2y + 5 = 0 , its equation is
The line passing through (1, 1) and parallel to the line 2x – 3y + 5 = 0 is
Correct option (A) 2x  3y + 1 = 0
Explanation:
Let the required line is 2x  3y + k = 0
it passes through (1, 1)
:. 2  3 + k = 0
k = 1
:. required line is 2x  3y + 1 = 0
The equation of the line which passes through the point (2 ,  3) and cuts off equal intercepts from the axis is
The vertex A of a triangle ABC is the point (2, 3) whereas the line perpendicular to the sides AB and AC are x – y – 4 = 0 and 2x – y – 5 = 0 respectively. The right bisectors of sides meet at P(3/2 , 5/2) . Then the equation of the median of side BC is
Given vertex of ΔABC A(−2,3)
Equation of side AB perpendiuclar to x−y−4=0 is
AB:x+y−λ=0
Side AB passes through point A(−2,3)
−2+3−λ=0
λ=1
Hence AB:x+y−1=0−−−−(1)
Equation of side AC perpendiuclar to 2x−y−5=0 is
AC:x+2y−λ=0
Side AC passes through point A(−2,3)
−2+6−λ=0
λ=4
Hence AC:x+2y−4=0−−−−(2)
The right bisectors of all sides are meeting at point P(3/2 , 5/2)
Right bisector from point B on side AC is perpendicular
Hence Equation of right bisector perpendicular to x+2y−4=0 is
2x−y−λ=0
Here right bisector is passing through point P (2 × 3/2 − 5/2)λ = 0
=> 6/2 − 5/2 = λ
λ=1/2
Equation of right bisector BD be
2x−y−1/2 =0
4x−2y−1=0−−−−(3)
Here On solving equation of AB and BD we get point of intersection i.e. point B
4(1−y)−2y−1=0
4−4y−2y−1=0
−6y=−3
y = 1/2
x = 1 − y = 1− 1/2 = 1/2
Point B(1/2,1/2)
Right bisector from point A on side BC is passing through point P
Hence equation of AD from point A and P
y−3 = {[5/2−3](x+2)/(3/2+2)}
y−3 = −1/7(x+2)
Slope of AD is mAD = −1/7
Hence slope of side BC perpendicular to AD is −1/7mBC = −1
mBC = 7
Equation of side BC from point B(1/2,1/2) with slope mBC=7
y − 1/2 = 7(x−1/2)
2y−1=14x−7
14x−2y−6=0
The distance between the lines 4x + 3y = 11 and 8x + 6y = 15 is ?
Given lines are 4x + 3y = 11 and 4x + 3y = 15/2
Distance between two parallel lines = c₁  c₂ / √a² + b²
=  11  15/2/ √16 + 9 
=  7 / 2 × 5 
= 7/10
The line which passes through the point (0 , 1) and perpendicular to the line x – 2y + 11 = 0 is
Thee perpendicular distance of the origin from the line 3x +4y + 1 = 0 is
Line y = 0 is Xaxis slope of this line = 0 = m1
line xy = 0 is y = x which is passes through (0,0) slope of this line = 45 = m2
tanФ = (m2m1)/(1m1m2) = (10)/(11*0)
= 1/1 = 1
tanФ =1
Ф = 45
The vertices of a triangle are (0 , 3) , ( 3 , 0) and (3 , 0). The orthocenter of the triangle is
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