![\"\"](\"/images/stars/stars-10.gif\")
You can put this solution on YOUR website!
(1)Find the Equation of a line through (0,10)
\n" ); document.write( "which is perpendicular to the line
\n" ); document.write( "-x-15y=10.\r
\n" ); document.write( "\n" ); document.write( "The given line is
\n" ); document.write( "That is -x-15y-10=0
\n" ); document.write( "That is x + 15y +10 =0 ----(1)
\n" ); document.write( "Any line perpendicular to (1) is given by
\n" ); document.write( "15x-y +k =0 ----(2)
\n" ); document.write( "Given that P(0,10) is a point on (2)
\n" ); document.write( "Therefore x = 0, y = 10 in (2)
\n" ); document.write( "15X(0) - 10 +k=0
\n" ); document.write( "0-10+k=0
\n" ); document.write( "k-10 = 0
\n" ); document.write( "k = 10 ----(*)
\n" ); document.write( "Putting (*) that is k=10 in (2)
\n" ); document.write( "15x-y +10 =0
\n" ); document.write( "Answer: The required line is 15x-y+10 =0\r
\n" ); document.write( "\n" ); document.write( "Note:(how do we consider (2) perpendicular to (1), looking at (1)?
\n" ); document.write( "Interchange the coefficients of x and y in (1) and (2) numerically and
\n" ); document.write( "change the sign of the coefficient of y in (2)(that is if it is negative in (1)
\n" ); document.write( "then positive in (2) and if positive in (1) then negative in (2))
\n" ); document.write( "Why should you do this ? How does it establish perpendicularity?
\n" ); document.write( "The principle is: when two lines are perpendicular,the product of their slopes is equal to (-1) That is if one line has slope = m, then a line perpendicular to it has slope =(-1/m)
\n" ); document.write( "In (1) we observe that the slope is (-1/15)
\n" ); document.write( "And therefore we should get slope of perpendicular line as 15
\n" ); document.write( "and hence the interchange of the coefficients(numerically)
\n" ); document.write( "and changing the sign of coefficient of y
\n" ); document.write( "How do you remember it in the form of a formula?
\n" ); document.write( "If (A)x+(B)y+C = 0 ----(1) is the given line
\n" ); document.write( "slope = (-A/B)----(*)
\n" ); document.write( "any line perpendicular to (1) is taken as
\n" ); document.write( "(Interchange the coefficients of x and y in (1) and (2) numerically and
\n" ); document.write( "change the sign of the coefficient of y in (2))
\n" ); document.write( "(B)x -(A)y +k = 0 ----(2)
\n" ); document.write( "slope = (B/A)-----(**)
\n" ); document.write( "From (*) and (**) the product of the slopes = (-A/B)X(B/A) = -1
\n" ); document.write( "Recall: that when two lines are parallel slopes are equal
\n" ); document.write( "and hence any line parallel to (1) should be of the form
\n" ); document.write( "(A)x+(B)y+k'= 0 ---(3)
\n" ); document.write( "slope= (-A/B)
\n" ); document.write( "The x and y-terms the same in both (1) and (3) \r
\n" ); document.write( "\n" ); document.write( "What more do you observe?
\n" ); document.write( "A very important observation is that (2) is not a single line but a family of lines with every member of the family perpendicular to (1). Giving different values to k we get different lines each perpendicular to (1)
\n" ); document.write( "Then how is it that we are asked to give the equation to a particular line perpendicular to (1)?
\n" ); document.write( "Every value of k determines a line perpendicular to (1)\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Problem Number 2)
\n" ); document.write( "2)The line segments with endpoints (0,5) and (5,0) is the diameter of a circle. write the eguation of this circle in standard form.\r
\n" ); document.write( "\n" ); document.write( "The equation to the circle with A(x1,y1) and B(x2,y2)
\n" ); document.write( "as the ends of a diameter of a circle is given by
\n" ); document.write( "(x-x1)(x-x2)+(y-y1)(y-y2) = 0
\n" ); document.write( "Therefore the equation to the circle with A(0,5) and B(5,0)
\n" ); document.write( "as the ends of a diameter of a circle is given by
\n" ); document.write( "(x-0)(x-5)+(y-5)(y-0) = 0 [here x1= 0,y1 = 5; x2 = 5, y2 = 0]
\n" ); document.write( "x(x-5)+(y-5)y = 0
\n" ); document.write( "x^2-5x+y^2-5y = 0
\n" ); document.write( "x^2+y^2-5x-5y = 0
\n" ); document.write( "Answer: The equation to the circle with (0,5) and (5,0)
\n" ); document.write( "as ends of a diameter is given by
\n" ); document.write( "x^2+y^2-5x-5y = 0\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Note: You can easily observe that A(5,0) is a point on the x-axis,distance 5 units from the origin and B(0,5) is a point on the y-axis, distance 5 units from the origin and since Ab is diameter and therefore angle AOB is the angle on the diameter =90(which we already know as the angle bet the x and y axes)
\n" ); document.write( "Therefore O(0,0) is a point on the circle. And that is the reason why the free constant in the answer for the equation to the circle is zero
\n" ); document.write( "The mid point of AB is given by
\n" ); document.write( "C=((0+5/2, 5+0/2)) = (5/2,5/2)
\n" ); document.write( "Distance OC is the radius
\n" ); document.write( "Now OC^2 = (5/2-0)^2 + (5/2-0)^2 = 25/4+25/4 = 50/4 = 25/2
\n" ); document.write( "Therefore radius = OC = sqrt(25/2) =5/(rt2)
\n" ); document.write( "The centeric C=(5/2,5/2)and radius 5/(rt2)
\n" ); document.write( "Therefore the general form of equation to the circle is
\n" ); document.write( "(x-h)^2+(y-k)^2 = r^2 where C=(h,k) and radius = r
\n" ); document.write( "Therefore our circle is
\n" ); document.write( "(x-5/2)^2+(y-5/2)^2 = 25/2
\n" ); document.write( "x^2+y^2-5x-5y=0 on simplification\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "