On L1, as x increases by 3, y increases by 5*3, so (4,17) is on L1
The perpendicular line has slope -1/5, so L2 is
y-17 = -1/5 (x-4)
Now you can convert that to slope-intercept form.
A line L1 passes through point (1, 2) and has gradient of 5. Another line L2, is perpendicular to L2 and meets it at a point where x = 4. Find the equation for L2 in the form of y = mx + c
5 answers
L1:
y-2 = 5(x-1)
y-2 = 5x-5
5x - y = 3
L2: x + 5y = c
from L1, when x = 4, y = 17
so (4,17) into L2
4 + 85 = c = 89
L2: x + 5y = 89 or y = (-1/5)x + 89/5
y-2 = 5(x-1)
y-2 = 5x-5
5x - y = 3
L2: x + 5y = c
from L1, when x = 4, y = 17
so (4,17) into L2
4 + 85 = c = 89
L2: x + 5y = 89 or y = (-1/5)x + 89/5
The line l1 has equation 3x + 5y − 7 = 0
(a) Find the gradient of l1
(2)
The line l2 is perpendicular to l1 and passes through the point (6, −2).
(b) Find the equation of l2 in the form y = mx + c, where m and c are constants.
(a) Find the gradient of l1
(2)
The line l2 is perpendicular to l1 and passes through the point (6, −2).
(b) Find the equation of l2 in the form y = mx + c, where m and c are constants.
The point A has coordinates (−4, 11) and the point B has coordinates (8, 2).
(a) Find the gradient of the line AB, giving your answer as a fully simplified fraction.
(2)
The point M is the midpoint of AB. The line l passes through M and is perpendicular to AB.
(b) Find an equation for l, giving your answer in the form px + qy + r = 0 where p, q and r are integers to be found.
(4)
The point C lies on l such that the area of triangle ABC is 37.5 square units.
(c) Find the two possible pairs of coordinates of point C.
(a) Find the gradient of the line AB, giving your answer as a fully simplified fraction.
(2)
The point M is the midpoint of AB. The line l passes through M and is perpendicular to AB.
(b) Find an equation for l, giving your answer in the form px + qy + r = 0 where p, q and r are integers to be found.
(4)
The point C lies on l such that the area of triangle ABC is 37.5 square units.
(c) Find the two possible pairs of coordinates of point C.
Gradient=(Y2-Y1)/(X2-X1)
5=(y-2)/(4-1)
y=17
gradient of L1×gradient of L2=-1
5×L2=-1
Gradient of L2=-1/5
5(y-17)=-1(x-1)
y=-1/5x+17 1/5
5=(y-2)/(4-1)
y=17
gradient of L1×gradient of L2=-1
5×L2=-1
Gradient of L2=-1/5
5(y-17)=-1(x-1)
y=-1/5x+17 1/5