Test: Dimensional Geometry - 7 - Mathematics MCQ

Proof: The parallel straight lines are
y = mx + c - ( i )
and y = mx + d ...(ii)
Let (h, k) be a point on (i)
∴ k = mh + c=> c = k - m h . ...(iii)
The perpendicular distance p from (h, k) on (ii) is given by

Comments about incentre.
1. It is the point of intersection of the internal bisectors of the angles.
2. It is equidistant from the sides
3. It is the centre of the incircle -touching all the sids of the triangle.
4. Its coordinates are

Proof: Let y = mx be die straight line common to ax² + 2hxy + by² = 0 ...(i)
and a'x² + 2h'xy + b'y² = 0 ...(ii)
So diat let the straight lines represented by (i) be 

The given equation which represents three equation is
ax³ + bx²y + cxy² + dy³ = 0
Since two of the straight lines are at right angles, therefore their equations can be taken as

The equation of the bisectors of the angles between the two straight lines given by ax²+2hxy+by²=0 is given by

Now we are given two pairs of straight lines, namely,
x²- 2 pxy - y² = 0 ...(I)
and x²- 2qxy-y² = 0 ...(II)
It is given that each of I or II is the bisectors of the angle between the other pair. Using (A) the equation of the bisectors of the angle between the straight lines represented by (I) is given by

Clearly the centre is (h, 0) and the radius is a.
∴ Equation of the circle is given by
(x - h )²+ (y -0 )² = er
 or x³ + y² - 2hx + h² = a²

Since the circle passes through the origin and has x-axis as its diameter, therefore its centre is [h, 0) and radius = h
∴ Required equation of the circle is
(x-h)² + (y-0)² = h²
or x² + y² -2hx = 0

Proof: If d is the required distance of the point (2, 3, 4) from the plane
3x -6y + 2z + 11 = 0
Then

To find the distance between two parallel planes, we follow the method given below. Method:Find the perpendicular distance of each plane from the origin with proper sign (i.e. do not. lake the mod). Ther. their differeee is the required distance between two parallel planes. Let d. and d, be the distances of the planes 
2x -2y + z + 3 =0 ...(i)
and 4x - 4y + 2z + 5 = 0 ...(ii)
from the origin. Then

Condition for the homogeneous second degr ee equation to represent two planes.
The most general homogeneous, second degree equation in x and y is
ax² + by² + cz² - 2fyx + 2gzx + 2hxy = 0 ...(1)
equation (i) will represent two p\anes (both passing through the origin) if

Let the two planes r epresented by




Condition of perpendicularity.
The two planes are perpendicular if
θ = 90° or tan θ = ∞
or a + b + c = 0
Remark: since the two planes pass through the origin, therefore the two planes are never parallel. The two planes can however coincide.

Planes represented by (b) are perpendicular planes.
Proof: The planes are given by
6x²+4y²-10z²+3yz+4zx-11xy=0
∴ a=6, b=4, c=-10
∴ a+b+c=6-4-10=0
Remark. Two planes arc perpendicular if the sum of the coefficients of x², y² and z² is zero.

Equations of planes bisecting the angle between two given planes.
Let the given planes 

If (x, y, z) be a point cm any one of the planes bisecting the angles between the planes (i) and (ii), then the perpendiculars from (x,z,x) on the two planes must be equal in magnitude (may be opposite in sign) so that

are the equations of the two bisecting planes.
Remark: 1- Sometimes we are required to find which of the planes bisect the angle between the given planes that contains the origin. To do this, we express the equations of the given planes in the form so that the constant terms in both of them are positive.

bisects that angle between the planes which contains the origin.
Naturally, the other bisecting plane

bisects the angle between the planes that does not contain the origin.
Remark: 2- If a₁a₂ + b₁b₂ + c₁c₂ < 0, then the origin lies in the acute angle and if a₁a₂ + b₁b₂ + c₁c₂ > 0 then the origin lies in the obtuse angle.

Proof: The given planes are
3x + 4y- 5z + 1 = 0 ...(i)
and 5x+12y -13z + 0 = 0 ...(ii)
Since the constant terms are already positive, therefore the equation of the bisector of the angle between (i) and (ii) that contains the origin is given by

= 15 - 48 4 65 > 0
therefore the origin lies in the obtuse angle. It follows that the plane
14x-8y + 13=0 is the equation of the bisector of the angle between the planes containing the origin and bisecting the obtuse angle

Note that two planes intersect in a line.

Through a given line infinitely many planes can pass. Any two such planes are intersecting in the given line. Hence infinitely many pairs of equations of first degree in x,y,z represent a straingt line.