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Question 30627: (1)Find the Equation of a line through (0,10) which is perpendicular to the line
-x-15y=10.
(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.
Found 2 solutions by sdmmadam@yahoo.com, venugopalramana:
Answer by sdmmadam@yahoo.com(530)   (Show Source):
You can put this solution on YOUR website! (1)Find the Equation of a line through (0,10)
which is perpendicular to the line
-x-15y=10.
The given line is
That is -x-15y-10=0
That is x + 15y +10 =0 ----(1)
Any line perpendicular to (1) is given by
15x-y +k =0 ----(2)
Given that P(0,10) is a point on (2)
Therefore x = 0, y = 10 in (2)
15X(0) - 10 +k=0
0-10+k=0
k-10 = 0
k = 10 ----(*)
Putting (*) that is k=10 in (2)
15x-y +10 =0
Answer: The required line is 15x-y+10 =0
Note:(how do we consider (2) perpendicular to (1), looking at (1)?
Interchange the coefficients of x and y in (1) and (2) numerically and
change the sign of the coefficient of y in (2)(that is if it is negative in (1)
then positive in (2) and if positive in (1) then negative in (2))
Why should you do this ? How does it establish perpendicularity?
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)
In (1) we observe that the slope is (-1/15)
And therefore we should get slope of perpendicular line as 15
and hence the interchange of the coefficients(numerically)
and changing the sign of coefficient of y
How do you remember it in the form of a formula?
If (A)x+(B)y+C = 0 ----(1) is the given line
slope = (-A/B)----(*)
any line perpendicular to (1) is taken as
(Interchange the coefficients of x and y in (1) and (2) numerically and
change the sign of the coefficient of y in (2))
(B)x -(A)y +k = 0 ----(2)
slope = (B/A)-----(**)
From (*) and (**) the product of the slopes = (-A/B)X(B/A) = -1
Recall: that when two lines are parallel slopes are equal
and hence any line parallel to (1) should be of the form
(A)x+(B)y+k'= 0 ---(3)
slope= (-A/B)
The x and y-terms the same in both (1) and (3)
What more do you observe?
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)
Then how is it that we are asked to give the equation to a particular line perpendicular to (1)?
Every value of k determines a line perpendicular to (1)
Problem Number 2)
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.
The equation to the circle with A(x1,y1) and B(x2,y2)
as the ends of a diameter of a circle is given by
(x-x1)(x-x2)+(y-y1)(y-y2) = 0
Therefore the equation to the circle with A(0,5) and B(5,0)
as the ends of a diameter of a circle is given by
(x-0)(x-5)+(y-5)(y-0) = 0 [here x1= 0,y1 = 5; x2 = 5, y2 = 0]
x(x-5)+(y-5)y = 0
x^2-5x+y^2-5y = 0
x^2+y^2-5x-5y = 0
Answer: The equation to the circle with (0,5) and (5,0)
as ends of a diameter is given by
x^2+y^2-5x-5y = 0
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)
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
The mid point of AB is given by
C=((0+5/2, 5+0/2)) = (5/2,5/2)
Distance OC is the radius
Now OC^2 = (5/2-0)^2 + (5/2-0)^2 = 25/4+25/4 = 50/4 = 25/2
Therefore radius = OC = sqrt(25/2) =5/(rt2)
The centeric C=(5/2,5/2)and radius 5/(rt2)
Therefore the general form of equation to the circle is
(x-h)^2+(y-k)^2 = r^2 where C=(h,k) and radius = r
Therefore our circle is
(x-5/2)^2+(y-5/2)^2 = 25/2
x^2+y^2-5x-5y=0 on simplification
Answer by venugopalramana(3286)  (Show Source):
You can put this solution on YOUR website! (1)Find the Equation of a line through (0,10) which is perpendicular to the line
-x-15y=10.
EQN. OF A LINE PERPENDICULAR TO THIS
15X-Y=K.......WHERE K IS A CONSTANT TO BE FOUND..THIS GOES THROUGH (0,10)..SO
15*0-10=K
K=-10
SO EQN. OF REQD.LINE IS
15X-Y=-10
15X-Y+10=0
(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.
TAKE P(X,Y) ANY POINT ON CIRCLE.THE 2LINES JOINING P TO ENDS OF DIAMETER ARE AT RIGHT ANGLES.HENCE PRODUCT OF THEIR SLOPES =-1....SO.....
EQN OF CIRCLE IS GIVEN BY
{(Y-5)/(X-0)}*{(Y-0)/X-5)}=-1
(Y-5)*Y/(X*(X-5))=-1
Y^2-5Y=-X^2+5X
X^2+Y^2-5X-5Y=0
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