Line L is given by the equation 3y+2x =9 and point P has coordinates (6,-5).
a. Explain why point P is not on the Line L
b. Find the gradient of line L
c. 1.) Write down the gradient of a line perpendicular to line L
2.) Find the equation of the line perpendicular to L and passing through point P.
For 2b ii, the answer would be different because the equation is asking for the line perpendicular to L passing through point P. The perpendicular slope is found to be 3/2, so the equation using point-slope form becomes:
y + 5 = (3/2)(x - 6)
Multiplying both sides by 2:
2y + 10 = 3x - 18
Rearranging:
3x - 2y + 28 = 0
Which is equivalent to:
2x + 3y + 3 = 0
So both answers are correct, just written in different forms.
For 2a explanation:
Since P(6,-5) does not satisfy the equation, L does not pass through point P.
On the other hand, 2b ii answer would be:
2x+3y+3=0
a. Well, point P is not on line L because when we substitute the x and y coordinates of P into the equation 3y + 2x = 9, it doesn't satisfy the equation. It's like trying to fit a square peg into a round hole - it just doesn't work!
b. To find the gradient of line L, let's rearrange the equation 3y + 2x = 9 into y = mx + c form, where m represents the gradient.
So, if we subtract 2x from both sides, we get 3y = -2x + 9.
Now divide both sides by 3, and we have y = (-2/3)x + 3.
The coefficient of x (-2/3) represents the gradient of line L.
c. 1.) The gradient of a line perpendicular to line L can be found by taking the negative reciprocal of the gradient of line L. So, if the gradient of line L is -2/3, the gradient of a line perpendicular to L would be the negative reciprocal, which is 3/2.
2.) To find the equation of the line perpendicular to L and passing through point P, we'll use the point-slope form of a line. This form is y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the gradient.
So substituting in the values for point P and the gradient we found (3/2), we get y - (-5) = (3/2)(x - 6).
Simplifying this, we have y + 5 = (3/2)x - 9.
Now, let's rearrange it in the form y = mx + c:
y = (3/2)x - 9 - 5,
which can be further simplified to y = (3/2)x - 14.
So, the equation of the line perpendicular to L and passing through point P is y = (3/2)x - 14.
a. To determine if point P is on line L, we can substitute the x and y coordinates of P into the equation of line L and see if it satisfies the equation. Let's check:
Substituting x=6 and y=-5 into the equation 3y + 2x = 9:
3(-5) + 2(6) = -15 + 12 = -3
Since -3 is not equal to 9, point P is not on line L.
b. The gradient (or slope) of a line can be found by rearranging the equation of the line to the form y = mx + c, where m is the gradient.
Rearranging the given equation 3y + 2x = 9 to this form:
3y = -2x + 9
y = (-2/3)x + 3
Comparing this equation with y = mx + c, we can see that the gradient m of line L is -2/3.
c. 1.) The gradient of a line perpendicular to another line is the negative reciprocal of the gradient of the original line. In this case, the gradient of line L is -2/3, so the gradient of the perpendicular line will be the negative reciprocal of -2/3.
The negative reciprocal of -2/3 is 3/2. So, the gradient of a line perpendicular to line L is 3/2.
c. 2.) To find the equation of the line perpendicular to line L and passing through point P, we can use the point-slope form of a line: y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the gradient.
Using point P(6, -5) and the gradient of 3/2, we have:
y - (-5) = (3/2)(x - 6)
y + 5 = (3/2)(x - 6)
y + 5 = (3/2)x - 9
y = (3/2)x - 14
Therefore, the equation of the line perpendicular to line L and passing through point P is y = (3/2)x - 14.
3(-5) + 2(6) = what??
-15 + 12 = -3 which is NOT 9
now
y = m x + b
3y+2x =9 ---> y = (-2/3) x + 3
slope = m = -2/3
perpendicular slope = m' = -1/m = 3/2
y = m' x + b
y = (3/2) x +b
at (6,-5)
-5 = (3/2)6 + b
b = -5 - 9 = -14
y = (3/2) x -14
2 y = 3 x - 28