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# l is the line x+2y-6=0. The line k is perpendicular to l and its y intercept is -7. Find the equation of k

(a) l is the line x+2y-6=0. The line k is perpendicular to l and its y intercept is -7. Find the equation of k

(b) m and n are 2 lines which pass trough the point of intersection of lines l and k (6,-3) is a point on m. Find the equation of m.

(ii) Find the two possible equations of n, if the angle between m and n is 45 degrees.

### 1 Answer by Expert Tutors

Vivian L. | Microsoft Word/Excel/Outlook, essay composition, math; I LOVE TO TEACHMicrosoft Word/Excel/Outlook, essay comp...
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Hi Nick;
(a)  The linear formula...
x+2y-6=0
is in Standard Format...
Ax+By-C=0, neither A nor B equal zero and A is greater than zero.
The slope is -A/B.
-(1/2)=-1/2
The slope of the line perpendicular to such is its negative inverse...
+2/1=2
The Slope-Intercept linear formula is...
y=mx+b
m is the slope.
b is the y-intercept, value of y when x=0.  It is provided as (0,-7).
y=2x-7
Standard form is...
0=2x-y-7
or
2x-y-7=0

(b) m and n are 2 lines which pass trough the point of intersection of lines l and k (6,-3) is a point on m. Find the equation of m.
The two equations are...
x+2y-6=0
2x-y-7=0
We need to find where these two lines intersection.  The easiest technique is elimination.  For this, either variable must have the same coefficient in both equations.  I choose y.  Its coefficient in the first equation is 2, and the second is -1.  Let's take the second equation.
2x-y-7=0
Let's multiply both sides by 2...
2(2x-y-7)=(2)(0)
4x-2y-14=0
Let's add this to the other equation...
4x-2y-14=0
x+2y-6=0
5x+0-20=0

5x-20=0
5x=20
x=4

Let's plug this into either equation to establish the value of y at the point of intersection.  I randomly select the first...
x+2y-6=0
4+2y-6=0
2y-2=0
2y=2
y=1

Let's plug both values into the second equation to verify results...
2x-y-7=0
[(2)(4)]-1-7=0
8-8=0
0=0
The point of intersection is (4,1).
The other point of the equation is (6,-3).
We must first establish slope.  This is the change-of-y divided by the change-of-x...
m=(y-y1)/(x-x1)
m=(1--3)/(4-6)
m=(1+3)/(4-6)
m=4/-2
m=-2

Point-slope formula is...
y-y1=m(x-x1)
We will use either point.  I randomly select the first, (4,1)...
y-1=-2(x-4)
y-1=-2x+8
y=-2x+9
Let's plug-in the other point, (6,-3), to verify the result...
-3=[(-2)(6)]+9
-3=-12+9
-3=-3

(ii) Find the two possible equations of n, if the angle between m and n is 45 degrees.
(4,1) is one point.  One line is on one side of the line, the other line on the other side.
The two lines of n are perpendicular to each other.
I will think about it.  I do not know the answer, at this time.