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The line joining (3, -1) and (2, 3)
Proof: We have
therefore the line joining (3, -1) and (2, 3) is perpendicular to the line joining (5, 2) and (9, 3).
The equation of the bisector of the angle between the lines 3x - 4y + 7 = 0 and 12x - 3y -8 =0 , in w hich the origin lines, is given by
Equations of Bisectors of the Angles between two non-parallel lines.
Let the straight lines
Ax+ By + C = 0 and
A'x + B'y + C = 0
be non-parallel. Then the bisectors are
1. if C > 0, C' > 0 and AA' + BB' > 0, then (i) is the equation of obtuse - angle bisector (so the equation (ii) gives the acute - angle bisector).
2. If C > 0, C' > 0 and AA' * BB' < 0, then (i) is the equation of acute - angle bisector (so (ii) gives the obtuse - angle bisector)
3. If C and C' are of the same sign (cither both positive or both negative), then (i) is the bisector of that angle (acute or obtuse) in which the origin lies.
In Problem 42, the given equations are
3x-4y + 7= 0 and
12x - 5y = 8 = 0
Since C and C' are of opposite sign, therefore the bisector of the angle between (iii) and (iv), in which the origin lies in given by
The line joining the origin to the point of intersection of 2x + 5y- 4 = 0 and 3x-2y+ 2 = 0 is given by
The given straight lines are
Their point of intersection is obtained on solving equations (i) and (ii) and is
The area of the triangle formed by the lines y -x = 0, y + x = 0, x - c = 0 is given by
The triangle is formed bv the straight lines
Their points of intersection are the vertices of the triangle and are (0, 0), (c, c) and (c, -c) The required area Δ is given by
A homogeneous equation of nth degree represents
A homogeneous equation of nth degree in x and y is a0xn + a1xn-1 + a2xn-2+ a3xn-3 .....+ an-1xyn-1+ anyn=0 and it represents at most n straight lines passing through the origin.
The angle between the straight lines represented by the equation Ax2 + 2Bxy + Cy2 = 0 is given by
The lines represented by the equation Ax2 + 2Bxy + Cy2 = 0 are real, coincident or imaginary according as B2 - AC is
The lines are represented by
The lines represented by the equation Ax2 + 2Bxy + Cy2 = 0 are perpendicular if
The angle φ between the lines is given by
The general equation of second degree ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 may represent a pair of straight lines if
The condition that the equation ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represents a pair of straight lines is that af2+bg2+ch2+2jgh-abc=0
The lines joining ihe origin to the points of intersection of 2x2 + 3xy - 4x + 1 = 0 and 3x + y - 1 = 0 are given by
The required equation of lines is obtained bymaking equation
The area of the triangle ABC formed by the intercepts OA a, OB = h, OC = con the coordinate axes respectively by a plane is given by
If Ax, Ay, Az be the areas of projection of an area A on the three coordinate planes then A2=Ax2 + Ay2 + Az2
we have to find the area A of triangle ABC Ax = the projection of A on yz plane
Ay = the projection of A on zx plane
Az = the projection of A on xy plane
∴ A2=Ax2 + Ay2 + Az2
If two pairs of opposite edges of a tetrahedron art; at right angles, then the third pair
Let the coordinates of the vertices O, A, B, C o f tetrahedron be (α, β, γ), (a, 0, 0), (0, b, 0) and (0. 0, c) respectively. Then it is given that
OA is perpendicular to BC
and OB is perpendicular to CA
To prove that
OC is perpendicular to AB
If the pairs of opposite edges of a tetrahedron OABC are at right angles, then
How many arbitrary constants are there in the equation of a plane?
The general equation of first degree in x, y, z namely
represents a plane, where at least one of A, B and C is non-zero.
Remark: 1. Let A ≠ 0. Then equation (i)
Therefore the general equation of a plane involves three independent (essential) arbitrary constant.
Hence if three independent conditions are given then the plane can be determined completely.
Remark : 2. If D = 0 then the plane will pass through the origin.
The equation of which of the following degree in x, y, z represents a plane
Which one of the following is the equation of the plane which meets the coordinale axes in A,B,C such that the centroid of the triangle ARC is (a,b,c)?
Let the coordinates of A, B and C be (a, 0, 0), (0, β, γ) and (0,0, y) respectively. Then the coordinates of the centroid of ΔABC is given by
In other words, the coordinates of A, B and C are
(3a, 0, O), (0, 3b, 0) and (0, 0, 3c) respectively.
Verify that the secoordinates of A, B and C satisfy equation in statement (c)
If l, m, n are the direction of the normal to the plane and p be the perpendicular distance from the origin on it, then the equation of the plane is of the type
OP is perpendicular on the plane ABC
∴ OP = p and d.c.'s of OP are [1, m, n]
Let Q(x, y, z) be an arbitrary point on the plane.
OP is the projection of OQ on OP.
⇒ OP = I(x- 0) + m(y- 0) + n(z- 0)
⇒ lx + my + nz = p
This is the required equation of the plane in normal form.
The equation of the plane passing through the point A(a, b, c) and perpendicular to OA (O being origin) is given by
Proof: OA is normal to the plane.
or ax + by + cz = a2 + b2 + c2 (statement(al))
Further, Let P(x, y, z) be any point on the plane.
Then OA is perpendicular to PA.
=> (x-a)a + (y- b)b+(z- c)c=0
(This is statement (b))
.*. both (a) and (b) are correct.
The equation of the plane passing through (2, -3, 1) and perpendicular to the line joining the points (3, 4. -1) and (2, -1, 5) is given by
Let P(x, y, z) be an arbitrary point on the plane. Then the line PA joining the points P(x, y, z) and A(2, -3, 1) lies on the plane and the d.r.’s of PA are
x - 2, y + 3, z- 1
Further, the direction ratios of the line joining points B(3, 4. - 1 ) and C(2, 1,5) are
3 - 2 , 4 - ( - 1 ),- 1 -5
i.e. 1, 5, -6
Since BC is perpendicular to the plane and hence is perpendicular to line PA, therefore
( x - 2) 1 + (y + 3) 5 + ( z - 1) (- 6 ) = 0
or x + 5y - 6z + 19 = 0
The equation of the plane through the points (2, 3, 1) and (-4. 5. 3) and parallel to X - axis is given by
The plane is parallel' to x-axis. Therefore, it is perpendicular to yzplane.
Hence its equation should be of the form
By Cz + D = 0
Since, the plane passes through (2, 3, 1) and (4, -5, 3), therefore.