I. B. Instructions: Find the equation of the line which passes through the given point parallel/ perpendicular to the given line. Show your solution and write the answer in a slope-intercept form on a separate sheet of paper.
1.)through (-4,3) and perpendicular to 2y-4x + 7 = 0
2.) through (1,5) and parallel to y = 6x-13 over 8
3.) through (1,5) and parallel to y- x = 5
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Answers
Answer:
1) y=2x+11 and y=-1/2x+1
2) y=6x-1 and y=-1/6x+31/6
3) y=x+4 and y=-x+6
Step-by-step explanation:
1) Parallel
Make the equation into form y=mx+c.
y=2x-3.5
The gradient is 2.
The new equation is y=2x+c.
Substitute the values in so now 3=-8+c.
c=11
The final equation is y=2x+11
Perpendicular
Make the equation into form y=mx+c.
y=2x-3.5
The gradient of the perpendicular is -1/2
The new equation is y=-1/2x+c
Substitute the values in so now 3=2+c
c=1
The final equation is y=-1/2x+1
2) Parallel
If y=6x-13/8 then the gradient is 6.
The new equation is y=6x+c.
Substitute the values in so now 5=6+c
c=-1
The final equation is y=6x-1
Perpendicular
If y=6x-13/8 then the gradient of the perpendicular is -1/6
y=-1/6x+c is the new equation.
Substitute the values in so now 5=-1/6+c
c=31/6
The new equation is y=-1/6x+31/6
3) Parallel
Make the equation into form y=mx+c.
y=x+5
The gradient is 1
y=x+c is the new equation.
Substitute the values in so now 5=1+c
c=4
y=x+4 is the new equation.
Perpendicular
Make the equation into form y=mx+c.
y=x+5
The perpendicular gradient is -1.
y=-x+c is the new equation.
Substitute the values in so now 5=-1+c
c=6
The final equation is y=-x+6
Hope this helps
Given: The properties of some lines are given which are parallel/ perpendicular to any given line.
To find: Equation of the lines
Explanation: 1- Line passing through (4,3) and perpendicular to 2y-4x + 7 = 0
The equation of line can be written in slope point form y-y1 = m( x-x1) where m is the slope and (x1,y1) is a point.
Slope of 2y-4x+7 is given by:
= - coefficient of x / coefficient of y
= -(-4) / 2
= 4/2
= 2
The slope of the line which is perpendicular to any given line whose slope is m is given by -1/m.
Therefore, slope
= -1/slope of 2y-4x+7
= -1/2
Using slope point form:
y-3 = (-1/2) (x-4)
=>2(y-3) = -1 (x-4)
=> 2y-6 = -x +4
=> x+2y = 10
The equation of the line is x+2y=10.
2-passing through (1,5) and parallel to y = 6x-13
The equation of line can be written in slope point form y-y1 = m( x-x1) where m is the slope and (x1,y1) is a point.
Slope of 6x-y-13= 0 is given by:
= - coefficient of x / coefficient of y
= -6 / -1
= 6/1
= 6
The slope of the line which is parallel to any given line whose slope is m is given by m.
Therefore, slope= 6
Using slope point form:
y-5 = 6(x-1)
=> y-5 = 6x-6
=> 6x-y = 1
The equation of the line is 6x-y=1.
3-passing through (1,5) and parallel to y- x = 5
The equation of line can be written in slope point form y-y1 = m( x-x1) where m is the slope and (x1,y1) is a point.
Slope of y-x= 5 is given by:
= - coefficient of x / coefficient of y
= -(-1) / 1
= 1/1
= 1
The slope of the line which is parallel to any given line whose slope is m is given by m.
Therefore, slope= 1
Using slope point form:
y-5 = 1(x-1)
=> y-5 = x-1
=> x-y= -4
The equation of the line is x-y= -4.