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Q : 18    If  $p$ is the length of perpendicular from the origin to the line whose intercepts on
the axes are $a$ and $b$,  then show that  $\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}$.

we know that intercept form of line is we know that In this problem On squaring both the sides  we will get Hence proved

Q : 17      In the triangle $ABC$ with vertices  $A(2,3)$$B(4,-1)$ and  $C(1,2)$ , find the equation and length of altitude from the vertex $A$.

Let suppose foot of perpendicular is  We can say that line passing through point   is perpendicular to line passing through point  Now, Slope of line passing through point  is ,  And Slope of line  passing through point  is ,  lines are perpendicular Therefore, Now,  equation of line passing through point  (2 ,3)  and slope with 1                      -(i) Now,...

Q : 16     If $p$ and $q$ are the lengths of perpendiculars from the origin to the lines    $x\cos \theta -y\sin \theta =k\cos 2\theta$  and  $x\sec \theta +y\hspace{1mm}cosec\hspace{1mm}\theta =k$ , respectively,  prove that  $p^2+4q^2=k^2$

Given equations of lines are      and   We can rewrite the equation  as Now, we know that  In equation   Similarly, in the equation    Now,                                                                                                                                                                                                     Hence proved

Q : 15      The perpendicular from the origin to the line  $y=mx+c$  meets it at the point  $(-1,2)$. Find the values of $m$ and $c$

We can say that line passing through point   is perpendicular to line  Now, The slope of the line  passing through the point  is ,  lines are perpendicular Therefore,                - (i) Now, the point  also lies on the line  Therefore, Therefore, the value of m and C is    respectively

Q : 14     Find the coordinates of the foot of perpendicular from the point  $(-1,3)$  to the line $3x-4y-16=0$.

Let suppose the foot of perpendicular is  We can say that line passing through the point   is perpendicular to the line  Now, The slope of the line  is ,  And The slope of the line  passing through the point is,  lines are perpendicular Therefore, Now, the point  also lies on the line  Therefore, On solving equation (i) and (ii) we will get Therefore,

Q : 13     Find the equation of the right bisector of the line segment joining the points  $(3,4)$ and   $(-1,2)$

Right bisector means perpendicular line which divides the line segment into two equal parts Now, lines are perpendicular which means their slopes are negative times inverse of each other Slope of line passing through points    and     is Therefore, Slope of bisector line is Now, let (h , k) be the point of intersection of two lines  It is given that point (h,k) divides the line segment...

Q : 12     Two lines passing through the point  $(2,3)$  intersects each other at an angle of  $60^{\circ}$. If slope of one line is $2$, find equation of the other line.

Let the slope of two lines are    respectively It is given the lines intersects each other at an angle of    and slope of the line is 2 Now, Now, the equation of line passing through point (2 ,3) and with slope    is                          -(i) Similarly, Now , equation of line passing through point (2 ,3) and with slope    is                               -(ii) Therefore,...

Q : 11        Prove that the line through the point  $(x_1,y_1)$  and parallel to the line   $Ax+By+C=0$   is   $A(x-x_1)+B(y-y_1)=0.$

It is given that line is parallel to the line   Therefore, their slopes are equal The slope of line   ,  Let the slope of other line be m Then, Now, the equation of the line passing through the point   and with slope   is Hence proved

Q : 10     The line through the points  $(h,3)$  and  $(4,1)$  intersects the line $7x-9y-19=0$  at right angle. Find the value of  $h$

Line passing through points ( h ,3) and (4 ,1) Therefore,Slope of the line is   This line intersects the line   at right angle Therefore, the Slope of both the lines are negative times inverse of each other  Slope of line  , Now, Therefore, the value of h is

Q : 9     Find angles between the lines  $\sqrt{3}x+y=1$  and   $x+\sqrt{3}y=1$.

Given equation of lines are     and    Slope of line  is,  And  Slope of line   is ,  Now, if   is the angle between the lines Then, Therefore, the angle between the lines is

Q : 8         Find equation of the line perpendicular to the line $x-7y+5=0$ and having  $x$intercept $3$.

It is given that line is  perpendicular to the line  we can rewrite it as Slope of line    ( m' ) =  Now,  The slope of the line is       Now, the equation of the line with  intercept   i.e. (3, 0) and  with slope -7 is

Q : 7      Find equation of the line parallel to the line  $3x-4y+2=0$  and passing through  the point $(-2,3)$

It is given that line is parallel to line   which implies that the slopes of both the lines are equal we can rewrite it as The slope of line   =   Now, the equation of the line passing through the point  and with slope  is Therefore, the equation of the line is

Q : 6     Find the distance between parallel lines

(ii)  $l(x+y)+p=0$ and $l(x+y)-r = 0$

Given equations of lines are  and  it is given that these lines are parallel Therefore, Now, Therefore, the distance between two lines is

Q : 6      Find the distance between parallel lines

(i) $15x+8y-34=0$  and  $15x+8y+31=0$.

Given equations of lines are    and   it is given that these lines are parallel Therefore, Now, Therefore, the distance between two lines is

Q : 5     Find the points on the x-axis, whose distances from the line  $\frac{x}{3}+\frac{y}{4}=1$  are  $4$  units.

Given equation of line is we can rewrite it as Now, we know that In this problem A = 4 , B = 3 C = -12 and d = 4 point is on x-axis therefore   = (x ,0) Now, Now if x > 3 Then,   Therefore, point is (8,0)            and if x < 3 Then, Therefore, point is (-2,0) Therefore, the points on the x-axis, whose distances from the line    are    units are  (8 , 0) and (-2 , 0)

Q : 4     Find the distance of the point  $(-1,1)$  from the line   $12(x+6)=5(y-2)$.

Given the equation of the line is we can rewrite it as Now, we know that         where A and B are the coefficients of x and y and  C is some constant  and   is point from which we need to find the distance  In this problem A = 12 , B = -5 , c = 82 and  = (-1 , 1) Therefore, Therefore, the distance of the point    from the line    is 5 units

Q : 3         Reduce the following equations into normal form. Find their perpendicular distances from the origin and angle between perpendicular and the positive x-axis.

(iii) $x-y=4$

Given equation is Coefficient of x is 1 and y is -1 Therefore,  Now, Divide both the sides by  we wiil get we can rewrite it as Now, we know that normal form of line is Where  is the angle between perpendicular and the positive x-axis and p is the perpendicular distance  from the origin On compairing equation (i) and (ii) we wiil get Therefore,  the angle between perpendicular...

Q : 3      Reduce the following equations into normal form. Find their perpendicular distances from the origin and angle between perpendicular and the positive x-axis.

(ii) $y-2=0$

Given equation is we can rewrite it as Coefficient of x is 0 and y is 1 Therefore,  Now, Divide both the sides by 1 we will get we can rewrite it as Now, we know that normal form of line is Where  is the angle between perpendicular and the positive x-axis and p is the perpendicular distance  from the origin On comparing equation (i) and (ii) we wiil get Therefore,  the...

Q : 3     Reduce the following equations into normal form. Find their perpendicular distances from the origin and angle between perpendicular and the positive x-axis.

(i)   $x-\sqrt{3}y+8=0$

Given equation is we can rewrite it as Coefficient of x is -1 and y is  Therefore,  Now, Divide both the sides by 2 we will get we can rewrite it as Now, we know that the normal form of the line is Where  is the angle between perpendicular and the positive x-axis and p is the perpendicular distance  from the origin On comparing equation (i) and (ii) we wiil get Therefore,  the...

Q: 2        Reduce the following equations into intercept form and find their intercepts on the axes.

(iii)  $3y+2=0$

Given equation is we can rewrite it as                           Therefore,  intercepts on y-axis are  and there is no intercept on x-axis
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