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klio [65]
3 years ago
12

Through (-3, 2) parallel to y = -2x-1

Mathematics
1 answer:
Slav-nsk [51]3 years ago
3 0

Answer:

y = - 2x - 4

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

y = - 2x - 1 ← is in slope- intercept form

with slope m = - 2

Parallel lines have equal slopes , then

y = - 2x + c ← is the partial equation

To find c substitute (- 3, 2 ) into the partial equation

2 = 6 + c ⇒ c = 2 - 6 = - 4

y = - 2x - 4 ← equation of parallel line

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Given: PS=RT, PQ=ST<br> Prove: QS=RS
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I) Eq(1) reason: sum of segments of a straight line

II) Eq(2) reason: Given PQ = ST & PS = RT

III) Eq(3) reason: sum of segments of a straight line

IV) Eq(4) reason: Same value on right hand sides of eq(2) and eq(3) demands that we must equate their respective left hand sides

V) Eq(5) reason: Usage of collection of like terms and subtraction provided this equation.

Step-by-step explanation:

We are given that;

PS = RT and that PQ = ST

Now, we want to prove that QS = RS.

From the diagram, we can see that from concept of sum of segments of a straight line we can deduce that;

PQ + QS = PS - - - - (eq 1)

Now, from earlier we saw that PQ = ST & PS = RT

Thus putting ST for PQ & PS for RT in eq 1,we have;

ST + QS = RT - - - - (eq 2)

Again, from the line diagram, we can see that from concept of sum of segments of a straight line we can deduce that;

RS + ST = RT - - - - -(eq 3)

From eq(2) & eq(3) we can see that both left hand sides is equal to RT.

Thus, we can equate both left hand sides with each other to give;

ST + QS = RS + ST - - - (eq 4)

Subtracting ST from both sides gives;

ST - ST + QS = RS + ST - ST

This gives;

QS = RS - - - - (eq 5)

Thus;

QS = RS

Proved

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