Test: Dimensional Geometry - 4 - Mathematics MCQ

The required equation
(i) makes an angle of 60° with x-axis
(ii) passes through the point (√3 , 2)
Now
(i) ⇒ equation of the straight line will be
y = tan 60° x + c
or y = √3 x + c ...(i)
(ii) ⇒ the coordinates of the point will satisfy equation (i)

First remember the following points:
l.The equation of a line passing through two p o in ts (x₁,y₁) and (x₂, y₂) is given by

2. The. coordinates of the point R which divides the line joining P(x₁,y₁) and Q(x₂, y₂) internally in the ratio m₁ : m₂ are given by

3. If R divides PQ externally in the ratio m₁ : m₂, then

Now the given straight line is
y - x + 2 - 0 ...(iv)
The equation of the straight line joining the points (3,-1) and (8, 9) is given by (see (i))

or y - 2x + 7 = 0 ...(v)
The point of intersection of (iv) and (v) is obtained on solving these equations and is given by (5,3)
Let the points of intersection (5, 3) divide the line joining (3. -4) and (8. 9) in the ratio m : n.
Then

 The given straight line is 

The equation of a straight line parallel to (i) will be

Since (ii) passes through P(4, 5), therefore the coordinates of P A\*ill satisfy equation (ii)

∴ Required equation of the straight line is

The condition for three lines lines to be concurrent is that

First of all learn the following:
Division of a plane by the line
L : Ax + By + C = 0
As shown in the adjoining Figure, the line L divides the plane into three parts, namely I, L and II, where
Ax₁ + By₁ + C > 0 for all poins (x₁, y₁) in l
Ax₂ + By₂ + C = 0 for all points (x₂, y₂) on L
Ax₃ + By₃ + C < 0 for all points (x₃, y₃) in II

Given point is (3, 2)
Given straight lines are

The given straight lines are

Comments. The angle φ between two straight lines
y = mx + c and y = m'x + c‘
is given by

Since the straight lines are ax + by + c = 0 and a'x +b'y + c' = 0
therefore

Both φ and π - φ shall be the angles between the given lines.

The angle φ between the lines is given by

Remark: A straight line passing through the origin shall be of the form
y = mx
and there will not bo any constant term in the eouation.

Proof in short : verify that
AB = 7 and AC = 3


where θ is the angle between AB (with d.c.'s [l₁, m₁, n₁]) and AC (with d.c.’s [l₂, m₂, n₂)
∴ the d.c.’s of the internal bisector are proprotional to l₁+l₂, m₁+m₂, n₁+n₂
or proportional to 
or proportional to 
or proportional to 25, 8, 5

In fact, we shall have

Proof: Let the three mutually perpendicular lines be

Proof: The area of the triangle OAB, where O is the origin and the coordinates of A and B are (x₁, y₁, z₁) a n d [x₂, y₂, z₂) respectively,

Here since the coordinates of A and B are (0,b,0) and (0.0,c), therefore the required area A is given by