Straight Lines - 1 - JEE MCQ

x₁ = (1*3 - 1*1)/(1-1)
y₁ = (1*4 - 1*2)/(1-1)
Hence the denominator becomes zero, here we can say that any point cannot be divided in the ratio of 1:1 externally.

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 (a²,ab) lie on the above line

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.

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