1) gradient of line = Δ y ÷ Δ x
= (5 -2) ÷ (3 - (-6))
= ¹/₃
using the point-slope form (y-y₁) = m(x-x₁)
using (3,5)
(y - 5) = ¹/₃ (x -3)
y - 5 = ¹/₃x - 1
⇒ <span> y = ¹/₃ x + 4 [OPTION D]
</span>2) y = 2x + 5 .... (1)
<span> </span>y = ¹/₂ x + 6 .... (2)
by substituting y in (1) for y in (2)
2x + 5 = ¹/₂ x + 6
³/₂ x = 1
x = ²/₃
by substituting found x (2)
y = ¹/₂ (²/₃) + 6
y = ¹⁹/₃
∴ common point is (²/₃ , ¹⁹/₃) thus answer is FALSE [OPTION B]
3) Yes [OPTION A]
This is because the both have a gradient of 5 and if they have the same gradient then that means that the two lines are parallel to each other.
4) No [OPTION B]
Two lines are perpendicular if their gradients multiply to give - 1 and as such one is the negative reciprocal of the other. Since both gradients are ¹/₂ then they are actually parallel and not perpendicular.
The answer would be 6.00^3.
Remember that .21 < .50
To do this, we will need to compare the slopes and the y int. So we will put each equation in y = mx + b form, where m will be the slope and b will be the y int..
4x + 5y = 10
5y = -4x + 10
y = -4/5x + 2.....the slope here is -4/5...the y int is 2
5x - 4y = 28
-4y = -5x + 28
y = 5/4x - 7...the slope here is 5/4...and the y int is -7
for the lines to be parallel, they would need the same slope and different y intercepts. So, as u can see, they are not parallel because they do not have the same slope.
for the lines to be perpendicular, the lines would have to have negative reciprocal slopes. They do....u have perpendicular lines. To find the negative reciprocal of a number, u flip the number and change the sign. One line had a slope of -4/5....the negative reciprocal is : flip the number making it -5/4....change the sign, making it 5/4...which is the slope of the other line...so these slopes are negative reciprocals.
so u have perpendicular lines
