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1.

Find the equations to the altitudes of the triangle whose angularpoints are `A(2,-2),B(1,1)a n dC(-1,0)dot`

Answer» `m=(y_2-y_1)/(x_2-x_1)`
`y-y_1=m(x-x_!)`
`m_(BC)=(0-1)/(-1-1)=1/2`
`m_(AD)=-2`
Equation of AD`y-(-2)=(-2)(x-2)`
`y+2=-2x+4`
`y+2x=2`
`A(2,-2),C(-1,0)`
`m_(AC)=(0-(-2))/(-1-2)=-2/3`
`m_(BE)=3/2`
Equation of BE`y-1=3/2(x-1)`
`(y-1)2=3x-3`
`2y-2-3x+3=0`
`2y-3x+1=0`
`A(2,-2),B(1,1)`
`m_(AB)=(1-(-2))/(1-2)=-3`
`C(-1,0)`
`y-0=1/3(x-(-1))`
`m_(CF)=1/3`
`3y=x+1`
`x-3y+1=0`.
Discussion
2.

Find the value of `m`for which the lines `m x+(2m+3)y+m+6=0a n dm x+(2m+1)x+(m-6)y+9=0`intersect at a; point on `y-a xi sdot`

Answer» `(0,(-m+6)/(2m+3))`
`(2m+3)y=-(m+6)`
`y=(-m+6)/(2m+3)`
`(m-6)y=-9`
`y=(-9)/(m-6)`
Point`(0,(-9)/(m-6))`
`(m+6)/(2m+3)=9/(m-6)`
`m^2-36=18m+27`
`m^2-18m-63=0`
`m^2-21m+3m-63=0`
`m(m-21)+3(m-21)=0`
`(m+3)(m-21)=0`
`m=21,-3`.
Discussion
3.

A line forms a triangle of area `54sqrt(3)s q u a r e`units with the coordinate axes. Find the equation of the line if theperpendicular drawn from the origin to the line makes an angle of `60^0`with the X-axis.

Answer» Equation of AB
`y=mx+c`
Equation of OD
`y-0=tan60(x-0)`
`u=sqrt3x`
`(m_(AB))sqrt3=-1`
`m_(AB)=-1/sqrt3`
Equation of line
`y=-1/sqrt3x+c`
`1/2*(sqrt3C)*C=54sqrt3`
`C^2/2=54`
`C^2=108`
`C=sqrt108`
`y=mx+c`
`y=(-1/sqrt3)x+sqrt108`.
Discussion
4.

Show that the line `x-y-6=0,4x-3y-20=0a d n6x+5y+8=0`are concurrent. Also, find their common point of intersection.

Answer» `x-y-6=0`
`x=y+6-(1)`
`4x-3y-20=0`
from equation 1
`4(y+6)-3y-20=0`
`y+4=0`
`y=-4`
putting this value in equation 1
`x=-4+6`
`x=2`
`=6(2)+5(-4)+8`
`=12-20+8=0`
Common point of intersection=[2,-4].
Discussion
5.

Find the equation of the internal bisector of angle `B A C`of the triangle `A B C`whose vertices `A ,B ,C`are`(5,2),(2,3)a n d(6,5)`respectively.

Answer» `(x_1,y_1),(x_2,y_2)`
`y-y_1=(y_2-y_1)/(x_2-x_1)(x-x_1)`
Equation of AC
`y-2=(5-2)/(6-5)(x-5)`
`y-2=3(x-5)`
`y-2=3x-15`
`3x-y-13=0`
Equation of AB
`y-2=(3-2)/(2-5)(x-5)`
`y-2=-1/3(x-5)`
`x+3y-11=0`
`|(a_1x+b_1y+c_1)/sqrt(a_1^2+b_1^2)|=pm|(a_2x+b_2y+c_2)/sqrt(a_2^2+b_2^2)|`
`3x-y-13=0`
`x+3y-11=0`
`(3x-y-13)/sqrt(9+1)=pmm(x+3y-11)/sqrt(1+9`
`3x-y-13=x+3y-11`
`2x-4y-2=0`
`3x-y-13=-x-3y+7`
`4x+2y-24=0`
`m=(-4)/2=-2`
`AB=x+3y-11=0`
`tantheta=|(m_1-m_2)/(1+m_1m_2)|=|((-1/3)-(-2))/(1+(-1/3)(-2))|`
`tantheta=((-1/3)+2)/(1+(2/3))`
`=(5/3)/(5/3)=1`
`tantheta=45^@`.
Discussion
6.

The equation of the base of an equilateral triangle is `x+y=2`and its vertex is `(2,-1)dot`Find the length and equations of its sides.

Answer» `tantheta=|(m_1-m_2)/(1+m_1m_2)|`
`theta=60,m_1=-1,m_2=?`
`tan60=|(-1-m_2)/(1-m_2)|`
`sqrt3=pm(-1-m_2)/(1-m_1)`
`sqrt3-sqrt3m_2=-1-m_2`
`m_2(1-sqrt3)=-1-sqrt3`
`m_2=(-1-sqrt3)/(1-sqrt3)`
`sqrt3-sqrt3m_2=1+m_2`
`m_2(1+sqrt3)=sqrt3-1`
`m_2=(sqrt3-1)/(1+sqrt3)`
`y-y_1=m(x-x_1)`
`y+1=(-1-sqrt3)/(1-sqrt3)(n-2)`
`y+1=(sqrt3+1)/(sqrt3-1)(n-2)`
`y+1=(sqrt3-1)/(sqrt3+1)(n-2)`.
Discussion
7.

Find the equation of a straight line which makes an angle of `tan^(-1)sqrt(2)`with the x-axisand cuts off an intercept of `=3/(sqrt(2))`with the y-axis

Answer» Equation of a straight line can be given as,
`y = mx+c`
Here, `y-` intercept `(c)= 3/sqrt2`
`theta = tan^-1(sqrt2)`
`=>tantheta = sqrt2`
`:. m = tan theta = sqrt2`
`:.` Equation of the line,
`y = sqrt2x+3/sqrt2`
`=>2x - sqrt2y +3 =0.`
Discussion
8.

Find the coordinates of the foot of the perpendicular drawn from thepoint `(1,-2)`on the line `y=2x+1.`

Answer» Here, slope of line, `y = 2x+1` is `2`.
`:.` Slope of lineperpendicular to this line `= -1/2`
Let foot of the perpendicular is `(x_1,y_1)`.
Then, `y_1 = 2x_1+1.`
So, from the point, `(1,-2)`,
`(2x_1+1+2)/(x_1-1) = -1/2`
`=>(2x_1+3)/(x_1-1) = -1/2`
`=>4x_1+6 = 1-x_1`
`=>x_1 = -1`
`:. y_1 = 2(-1)+1 = -1`
So, foot of the perpenicular will be `(-1-1).`
Discussion
9.

Find the equation of the straight line which cuts off intercept onX-axis which is twice that on Y-axis and is at a unit distance from theorigin.

Answer» Eqaution of a straight line can be given as,
`x/a+y/b = 1`
It is given that, `a = 2b`
`:. x/(2b) +y/b = 1`
Also, it is given that distance of this line from the origin is unit that is `1`.
`:. |sqrt(1/b^2+1/(4b^2))| = 1`
`=>1/b^2+1/(4b^2) = 1^2`
`=>5 = 4b^2`
`=>b = +-sqrt5/2`
Putting value of `b` in equation of the line,
`x/sqrt5+(2y)/sqrt5 = 1 and -x/sqrt5-(2y)/sqrt5 = 1` are the required equations.
Discussion
10.

Find the coordinates of the incentre and centroid of the triangle whosesides have the equations `3x-4y=0,12 y+5x=0 and y-15=0.`

Answer» 3x=4y
3x=4*15
x=20
x=-36
A(0,0),B(20,15).C(-36,15)
Centroid`=((0+20+(-36))/3,(0+15+15)/3)`
`=(-16/3,10)`
`a=sqrt((20+36)^2+0^2)=sqrt((56)^2)=56`
`b=sqrt(36^2+15^2)=39`
`c=sqrt(20^2+15^2)=25`
Discussion
11.

A straight line cuts intercepts from the axes of coordinates the sum of whosereciprocals is a constant. Show that it always passes though as fixed point.

Answer» `x/a+y/b=1`
`1/a+1/b=alpha`
`1/a=alpha-1/b`
`(alpha-1/b)x+y/b=1`
`1/b(y-alpha)+(alphax-1)=0`
`alpha^2-1=0`
`L_1=y=lambda=0`
`y=1/lambda(11/alpha,1/alpha)`.
Discussion