Lycée Jean-Piaget - Neuchâtel

Mathematics 2MPES Chapter 2

Parallel&PerpendicularLines

Exercise2.14

g. Graph 𝐿!, 𝐿!, 𝐿!

Parallel line passing through a given point. Let us consider the straight-line 𝐿! defined by the standard equation 5𝑥 + 8𝑦 = 17.

a. Put down the slope intercept form of 𝐿!:

𝐿!:𝑦! = − !!𝑥 + !"

!

b. What is the slope and y-intercept of 𝐿!?

Slope: 𝑚 = − !! y-intercept: !"

!

c. Determine the slope intercept form of 𝐿!, a straight-line parallel to 𝐿! if we call ℎ, the value of the y-intercept of 𝐿!.

𝐿!:𝑦! = − !!𝑥 + ℎ

d. What is the value of ℎ if the line 𝐿! passes through the point 𝑃!(8;−3)?

If 𝐿! passes through 𝑃!, then −3 = − !!8 + ℎ, so ℎ = 2

e. Determine the slope intercept form of 𝐿!, a straight-line parallel to 𝐿!,if we call 𝑘, the value of the y-intercept of 𝐿!.

𝐿!:𝑦! = − !!𝑥 + 𝑘

f. What is the value of 𝑘 if the line 𝐿! passes through the point 𝑃!(4;−!!)?

If 𝐿! passes through 𝑃!, then − !!= − !

!4 + 𝑘, so 𝑘 = 0

Conclusion: (Summarize your work by interpreting, transcribing and producing descriptions using your answers above, and the dedicated words of the green post-it.) Always put down in slope intercept form to evaluate the slope and y-intercept. Parallel lines share the same slope. To find the y-intercept of a parallel line that passes through a given point, substitute the coordinates of that point in the slope intercept form of the equation.

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Lycée Jean-Piaget - Neuchâtel

Mathematics 2MPES Chapter 2

Exercise2.15

g. Graph 𝐿!, 𝐿!, 𝐿!

y-intercept and x-intercept. Let us consider the straight-line 𝐿! defined by the standard equation 2𝑥 − 𝑦 = 1.

a. Put down the slope-intercept form of 𝐿!: 𝐿!: 𝑦! = 2𝑥 − 1 :

b. What value of 𝑥 do you choose to find the y-intercept of 𝐿!? Value of 𝑥: 𝑥 = 0 value of y-intercept: −1

c. What value of 𝑦 do you choose to find the x-intercept of 𝐿!?

Value of 𝑦: 𝑦 = 0 value of x-intercept: !!

d. Sketch the graph of the line on the left. At point 𝑃!(0;−1), sketch in red the slope triangle. Then rotate the three sides of the slope triangle by 90° counter-clockwise, and sketch this new triangle in blue. What is the value of the slope of the blue triangle?

𝑚 = ∆!∆!= !!

!so 𝑚 = − !

!

e. Determine the slope-intercept form of 𝐿!, a straight-line perpendicular to 𝐿! corresponding to the blue slope triangle.

𝐿!:𝑦! = − !!𝑥 − 1

f. What is the slope-intercept form of 𝐿! a straight-line perpendicular to 𝐿! that passes through the point 𝑃!(−1; 2)?

If 𝐿!: 𝑦! = − !!𝑥 + ℎ passes through 𝑃!, then 2 = − !

!−1 + ℎ, so ℎ = !

!

Conclusion: Always put down in slope intercept form to evaluate the slope and y-intercept. If Parallel lines share the same slope, with perpendicular lines, the product of the slopes is equal to “-1”. In other words, the slopes of two perpendicular lines are opposite reciprocals. To find the y-intercept of a perpendicular line that passes through a given point, substitute the coordinates of that point in the slope intercept form of the equation.

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Lycée Jean-Piaget - Neuchâtel

Mathematics 2MPES Chapter 2

Exercise2.16

g. Graph 𝐿!, 𝐿!, 𝐿!

Lines intersection. Let us consider the straight-line 𝐿! defined by the standard equation 2𝑥 − 𝑦 = −3 and consider the straight-line 𝐿! defined by the standard equation 𝑥 + 𝑦 = −3

f. Put down the slope-intercept form of 𝐿!: 𝐿!:𝑦! = 2𝑥 + 3:

g. What is the slope and y-intercept of 𝐿!? Slope: 𝑚 = 2 y-intercept: 3

h. Put down the slope-intercept form of 𝐿!: 𝐿!:𝑦! = −𝑥 − 3:

i. What is the slope and y-intercept of 𝐿!? Slope: 𝑚 = −1 y-intercept: −3

j. Verify that the point 𝑀(−2;−1) is on 𝐿! by substituting its coordinates in the equation of 𝐿!. If 𝑀(−2;−1) is on 𝐿!:𝑦! = 2𝑥 + 3, we have. −1 = 2 −2 + 3 correct!

k. Verify that the point 𝑀(−2;−1) is on 𝐿! by substituting its coordinates in the equation of 𝐿!.

If 𝑀(−2;−1) is on 𝐿!:𝑦! = −𝑥 − 3, we have.−1 = − −2 − 3 also correct! l. What is the property the point 𝑀 coordinates with respect to the slope-

intercept forms of 𝐿! and 𝐿!. At point 𝑀,𝑦! = 𝑦! so we can find x coordinate of the intersection by solving: 2𝑥 + 3 = −𝑥 − 3 , which gives 3𝑥 = −6 𝑠𝑜 𝑥 = −2

Conclusion: To find the coordinates of the intersection point between to lines, the same x-value gives the same y-value in both slope intercept equations. We have to equal the right members of the equation (𝑚!𝑥 + ℎ! = 𝑚!𝑥 + ℎ!) to find the x-coordinate, then we substitute this value in any of the equatin to get the y-coordinate.

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