1 Crore+ students have signed up on EduRev. Have you? 
Consider a family of straight lines (x + y) + λ(2x  y + 1) = 0 . Find the equation of the straight line belonging to this family that is farthest from (1,3)
AT FIRST FIND THE COMMON PONT OF THIS FAMILY WHICH IS A (1/3 , 1/3)
THEN FOR THE LINE TO BE AT THE FARTHEST DISTANCE FROM B(1,3) THE LINE SHOULD BE PREPENDICULAR TO AB THEREFORE THE SLOPE OF LINE m= 2/5
EQUATION OF LINE IS 2x + 2/3 = 5y  5/3 WHICH IS HENCE 6x15y7=0
One side of an equilateral triangle is 3x+4y=7 and its vertex is (1,2). Then the length of the side of the triangle is
Find the equation of the bisector of the angle between the lines x+2y–11 = 0, 3x–6y–5 = 0 which contains the point (1,–3)
x+2y−11=0 ..(1)
3x−6y−5=0 ..(2)
Point of intersection of (1) and (2)
x=19/3
y=7/3
Equation passing through (19/3,7/3) and (1,3)
y7/3=1616(x19/3)
xy4=0
A straight line which make equal intercepts on +ve x and y axes and which is at a distance '1' unit from the origin intersects the straight line y = 2x + 3 + √2 at (x_{0}, y_{0}) then 2x_{0} + y_{0}
Equation of the straight line having equal intercepts is x + y = k and proceed.
The locus of the image of the point (2, 3) in the line (x 2y +3) + λ (2x 3y + 4) = 0 (λ∈R) is
A(2,3),B(1,2),P(x,y)
AB = BP
A straight line L with negative slope passes through the point (8, 2) and cuts the positive coordinate axes at points P and Q. As L varies, the absolute minimum value of OP+OQ is (O is origin)
The equation of the line L be x coordinates of P and Q are 3x + 4y = 9 and y = mx + 1 and P (3, 4) .
So Q (0, 1)
So, absolute minimum value of x
The number of integral values of m for which the x coordinate of the point of intersection of the lines 3x + 4y = 9 and y = mx + 1 is also an integer is
Solving given equations we get is an integer, if
(2, 0) y = mx So m has two integral values.
Area of the parallelogram formed by the lines y = mx, y = mx+ 1, y = nx and y = nx+ 1 equals (m ≠ n)
If be the distance between parallel sides and θ be the angle between adjacent sides, then
Requried area Whre y = mx + 1, x (distance between   lines) and π/4 (0,3)
If the midpoint of sides of a triangle are (a, 0), (0, b) and (a, b), then the centroid of the triangle is
The triangle ABC has medians AD, BE, CF. AD lies along the line y = x + 3, BE lies along the line y = 2x + 4, AB has length 60 and angle C = 90°, then the area of ΔABC is
Let A(1, 0), B(0, –2) and P(a, b) be a third point that lies on the straight line 'x  y= 0' such that, PA + PB is minimum then the values of a + b =
Let A(0, 0), B(3, 0), C(2, 1) then area of the square inscribed in the triangle such that, one side of the square is taken along
Let O(0, 0), A(–1, 1) the point ‘A’ is rotated 60^{o} in anticlockwise sense about origin then the slope of line joining original and the new positions of ‘A’ is
Let y = 2x + c_{1}, y = 2x + c_{2} meet Xaxis at A_{1}, A_{2} respectively and Yaxis at B_{1}, B_{2} respectively then the locus of the point of intersection of lines
Let A(2, 3) and B is the image of A with respect to the lines (2x  3y + 4) + k (x  2y + 3) = 0 then the locus of B is
The given line is
(2x  3y + 4) + k (x  2y + 3) = 0
as x−2y=−3
2x−3y=−4
multiplying eqn(2) by 2 and subtracting we get, y=2
so, x=−3+4=1
so point of intersection P is (1,2)
A′ is the reflection of A
as, AP=BP=A′P
so, √(21)^{2}+(32)^{2}] =[(h1)^{2}+(k2)^{2}]^{1/2}
1+1=h^{2}+k^{2}2h4k+1+4
h^{2}+k^{2}−2h−4k+3=0
x^{2}+y^{2}2x4y+3=0
Let 'P' be a point inside the equilateral triangle ABC, PA = 3, PB = 4, PC = 5 then the area of the triangle ABC (in square units)
Rotate the triangle by 60^{o}
AC^{2} = AP^{2} + PC^{2}  2AP.PCcos(θ + 60°)
Let a straight line passing through P(1, 4) with negative slope cuts the coordinate axes at A, B then the area of the triangle OAB when OA + OB is minimum is
OA + OB = 1 + 4cot θ + 4 + tan θ
= 5 + tan θ + 4 cot θ
Attains minima only if tan θ = 4 cot θ
The image of the curve with respect to the straight line x + y= 1 is
Replace x by 1 – Y
y by 1 – X
Let (a, b) lies on obtuse angle bisector of x + 2y + 1 = 0, 2x + y + 1 = 0 then the least value of (a  1)^{2} + (b  2)^{2} is
Find short dist. From (1, 2)
Let A(0, 0) and x + y  1 = 0, 2x + 1 = 0 are the altitude, median from the vertices B, C respectively of ΔABC then the area of the triangle
Let (x  1)^{2} + (y  2)^{2} = 1/4 then the minimum value of " x + 3y+ 4"
Two vertical poles 20 mts, and 80 mts stand apart on a horizontal plane. The height of the point of intersection of the lines joining the top of each pole to the foot of others (in metres)
The height "a" of the point E of intersection of the lines AD & BC drawn from the tops B & D of two poles AB & CD , to the feet of the other pole is given by LADDER THEOREM .
1/a = 1/AB + 1/CD
1/a = 1/20 + 1/80
= 5/80 = 1/16
So. a = 16 m
A(3, 1), B(13, 6), C(13, 21), D(3, 16) forms a parallelogram, then the yintercept of the line passing through (1, 0) and dividing the parallelogram into two equal areas
If the straight lines ax + by + p = 0 and x cosa + y sina = p are inclined at an angle p/4 and concurrent with the straight line x sina – y cosa = 0, then the value of a^{2} +b^{2} is
ax+by+p=0 ...(1)
⇒by=−ax+p
⇒y=−a/bx+p/b
So, Slope,m1 = −a/b
and xcosθ+ysinθ−p=0 ..(2)
⇒ysinθ=−xcosθ+p
⇒y=−cosθ/sinθx+p/sinθ
So, Slope,m_{2}=−cosθ/sinθ
Now, straight lines (1) and (2) include an angle π/4
i,e tanπ/4 = ∣∣(m_{1}−m_{2})/(1+m_{1}m_{2})∣∣
⇒ 1 = ∣∣(m_{1}−m_{2})/(1+m_{1}m_{2})∣∣
⇒1+m_{1}m_{2}=m_{1}−m_{2}
⇒∣∣1+(−a/b×(−cosθ/sinθ))∣∣
=∣∣−a/b−(−cosθ/sinθ)∣∣
⇒∣∣(bsinθ+acosθ)/bsinθ∣∣
=∣∣(−asinθ+bcosθ)/bsinθ∣∣
⇒bsinθ+acosθ=−asinθ+bcosθ
Squaring both sides, we get
b^{2}sin^{2}θ+a^{2}cos^{2}θ+2absinθ cosθ=a2sin2θ+b2cos2θ−2absinθ cosθ ...(3)
Given three lines are concurrent
Now, we will find the point of intersection of
xcosθ+ysinθ=p and xsinθ−ycosθ=0
which is x=pcosθ and y=psinθNow, point of intersection will satisfy (1)
i.e apcosθ+bpsinθ+p=0
⇒apcosθ+bpsinθ=−p
⇒acosθ+bsinθ=−1
Squaring both sides, we get
⇒a^{2}cos^{2}θ+b2sin^{2}θ+2absinθ cosθ=1 ...(4)
Put the value of (4) in (3), we get
a^{2}sin^{2}θ+b^{2}cos^{2}θ−2absinθ cosθ =1 ....(5)
Adding (4) and (5), we get
a^{2}cos^{2}θ+b^{2}sin^{2}θ+2absinθ cosθ+a^{2}sin2θ+b^{2}cos^{2}θ−2absinθ cosθ=1+1
⇒a^{2}(sin^{2}θ+cos^{2}θ)+b2(sin^{2}θ+cos^{2}θ)=2
⇒a^{2}+b^{2}=2
Let (1, 2), (3, 8), (4, 1) are three vertices of a parallelogram the sum of all possible x coordinates of the fourth vertex of the parallelogram is
Let ABCD be a parallelogram.
then A (1,2), B (3,8) and C (4,1). Let the coordinates of D be (x,y)
Since the diagonals of a parallelogram bisect each other
Coordinates of the midpoint of AC = Coordinates of the midpoint of BD
(1+4)/2 , (2+1)/2 = (3+x)/2 , (8+y)/2
5/2 , 3/2 (3+x)/2 , (8+y)/2
2.5, 1.5 , (3+x)/2 , (8+y)/2
(3+x)/2 = 2.5 , 3/2 = (8+y)/2
(3+x) = 5 , 5 = y
2 = x , 5 = y
∴ Coordinates of the fourth vertex are (2, 5)
The orthocenter of the triangle formed by the lines x  7y + 6 = 0, 2x  5y  6 = 0 and 7x + y  8 = 0 is
x7y+6 = 0
m1 = 1/7 = 1/7
7x+y8 = 0
m2 = 7/1
Slope = 1/7 * (7)
= 1
B is orthocentre of triangle ABC
x7y = 6 (1)
7x+y = 8 (2)
y =87x (3)
Put eq (3) in eq(1)
x7(87x) = 6
x56+49x = 6
50x = 50
x = 1
Similarly y = 1
Points are (1,1)
The equations x = t^{3} + 9 and represents a straight line where t is a parameter. Then yintercept of the line is :
x=0
Put into x= t^{3} + 9
0 = t^{3} + 9
t^{3} = 9
y = 3(t^{3})/4 + 6
y = 3(9)/4 + 6
y= ¾
The vertices of triangle ABC are A(–1,–7), B(5, 1) and C(1, 4). The equation of the bisector of the angle ABC of ΔABC is :
y  1 = m_{1} x  3 and y  3 = m_{2} (x  1) are two family straight lines (m_{1}, m_{2}R), at right angles to each other. The locus of their point of intersection is :
Locus of P (h, k) is circle with AB as diameter
(x 3)(x 1) + (y 1)(y 3) = 0
The equation of the line passing through the intersection of the lines 3x + 4y = 5, 4x + 6y = 6 and perpendicular to 7x  5y + 3 = 0 is :
Equation of altitude on BC
x + 4y= 13
Equation of altitude on AB
7x  7y + 19 = 0
⇒
130 videos359 docs306 tests

Use Code STAYHOME200 and get INR 200 additional OFF

Use Coupon Code 
130 videos359 docs306 tests









